Answer:
Given : A bus of mass 760 kg requires 120 m to reach certain velocity value Vf.
the bus engine exerts a constant forward force F.
To Find : When the bus is towing a 330-kg small car, how long distance needed to reach same Vf?
Solution:
V² - U² = 2aS
V = Vf
U = 0
S = 120 m
=> Vf² - 0 = 2a(120)
=> Vf² = 240a
m = 760 kg
Force = F
F = ma
=> F =760 a
=> a = F/760
Vf² = 240F/760
Case 2 :When the bus is towing a 330-kg small car,
m = 760 + 330 = 1090 kg
a = F/1090
Vf² = 2aS
=> 240F/760 = 2 (F/1090) S
=> S = 120 x 1090 /760
=> S = 172.1 m
172.1 m distance needed to reach same Vf
Explanation:
A ball is thrown vertically upward from the top of a 100 foot tower, with an initial velocity of 10 ft/sec. Its position function is s(t)=−16t2+10t+100.
a. What is its velocity in ft/sec when t = 2 seconds? (Solve by using instantaneous rate.)
b. Determine the equation of a line, in slope-intercept form, that passes through the points (5, 6) and (10, 2).
Answer;
-54ft/s
y = -4/5 x + 10
Explanation
Given the position of an object expressed by the function
s(t)=−16t²+10t+100
Velocity is the change in position with respect to time
v(t) = ds(t)/dt
v(t) = -32t + 10
When t = 2
v(2) = -32(2)+10
v(2) = -64+10
v(2) = -54
Hence the velocity of the object is -54ft/s
b) The standard equation of a line in point slope form is expressed as;
y = MX+c
M is the slope
c is the y-intercept
Given the coordinate (5, 6) and (10, 2)
M = 2-6/10-5
M = -4/5
Get the y-intercept
Substitute m = -4/5 and any point say (5,6) into the expression y = mx+c
6 = -4/5 (5) + c
6 = -4+c
c = 6+4
c = 10
Get the required equation
Recall that: y = mx+c
y = -4/5 x + 10
Hence the equation of a line, in slope-intercept form is y = -4/5 x + 10
A charge of 7.1 x 10-4 C is placed at the origin of a Cartesian coordinate system. A second charge of 6.5 x 10-4 C lies 20 cm above the origin, and a third charge of 8.9 x 10-4 C lies 20 cm to the right of the origin. Determine the direction of the total force on the first charge at the origin. Express your answer as a positive angle in degrees measured counter clockwise from the positive x-axis.
Answer:
α = 36.21 °
β = 143.79°
Explanation:
To do this, we need to know the expression to calculate the angle.
In this case:
α₁ = tan⁻¹ (Fy₁/Fx₁) (1)
Now, let's analize the given data.
We have a charge q₁ at the origin of the cartesian coordinate system, so, it's at the 0. The charge q₂ is 20 cm above q₁, meaning is on the y-axis. Finally q₃ it's 20 cm to the right, meaning it's on the x-axis.
Knowing this,we can calculate the force that q₂ and q₃ are exerting over q₁. As these forces are in the x and y-axis respectively, we also are calculating the value of the forces in the x and y axis, that are needed to calculate the direction.
The expression to calculate the force would be Coulomb's law so:
F = K q₁q₂ / r² (2)
The value of K is 9x10⁹ N m² / C². Let's calculate the forces:
F₁₂ = Fy = 9x10⁹ * (7.1x10⁻⁴) * (6.5x10⁻⁴) / (0.020)²
Fy = 1.04x10⁷ N
F₁₃ = Fx = 9x10⁹ * (7.1x10⁻⁴) * (8.9x10⁻⁴) / (0.020)²
Fx = 1.42x10⁷ N
Now that we have both forces, we can calculate the magnitude of the force:
F = √(Fx)² + (Fy)²
F = √(1.04x10⁷)² + (1.42x10⁷)²
F = 1.76x10⁷ N
Finally, the direction would be applying (1):
α = tan⁻¹ (1.04x10⁷/1.42x10⁷)
α = 36.21 °
And counter clockwise it would be:
β = 180 - 36.21 = 143.79°
Hope this helps
calculate the magnitude and the direction of the resultant forces
answer:
resultant = 127.65 in the positive direction
explanation:
F1 = 50N , F2 = 40N, f3 = 55N , f4 = 60N
Fy = 50 sin 50 = 50 × -0.26 = -13
Fx = 40 cos 0 = 40×1 = 40
fx = 55 cos 25 = 55×0.99 = 54.45
Fy = 60 sin 70 = 60 × 0.77 = 46.2
resultant = -13+40+54.45+46.2 = 127.65 in the positive direction
Suppose it takes a constant force a time of 6.0 seconds to slow a 2500 kg truck
from 26.0 m/sec to 18.0 m/sec. What is the magnitude of the force? Give
your answer in scientific notation rounded correctly.
Answer:
[tex]3.3\cdot 10^3\:\mathrm{N}[/tex]
Explanation:
Impulse on an object is given by [tex]\mathrm{[impulse]}=F\Delta t[/tex].
However, it's also given as change in momentum (impulse-momentum theorem).
Therefore, we can set the change in momentum equal to the former formula for impulse:
[tex]\Delta p=F\Delta t[/tex].
Momentum is given by [tex]p=mv[/tex]. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:
[tex]p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}[/tex].
Now we plug in our values and solve:
[tex]\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}[/tex](two significant figures).
Calculate the approximate value for the distance travelled in the first 10 s by using the above graph
Answer:
80 m
Explanation:
From the question given above, the following data were obtained:
Speed = 8 m/s
Time = 10 s
Distance =?
Speed can be defined as the distance travelled per unit time. Mathematically, it can be expressed as:
Speed = Distance / time
With the above formula, we can obtain the distance travelled as illustrated below:
Speed = 8 m/s
Time = 10 s
Distance =?
Speed = Distance / time
8 = Distance / 10
Cross multiply
Distance = 8 × 10
Distance = 80 m
Therefore, the distance travelled is 80 m
cal ulate a moment of force o. 50 meter distance and 10 newton force
Answer:
500N/M
Explanation:
given that
force=10N
distance=50M
moment=force*distance
=10×50=500j
i’ll mark you brainlist :)
Two 0.60-kilogram objects are connected by a thread that passes over a light, frictionless pulley. The objects are initially held at rest. If a third object with a mass of 0.30 kilogram is added on top of one of the 0.60-kilogram objects and the objects are released, the magnitude of the acceleration of the 0.30-kilogram object is most nearly:______
Answer:
2 m/s²
Explanation:
From the given information:
The first mass m_1 = 0.6 kg
The second mass m_2 = 0.3 kg
The magnitude for the acceleration of 0.3 kg is:
a = net force/ effective mass
Mathematically, it can be computed as follows:
[tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)[/tex]
[tex]a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)[/tex]
a ≅ 2 m/s²
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 34.5 N is applied tangent to the rim of the disk.
a) What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through .200 revolution?
b) What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through .200 evolution?
Answer:
a) v = 1.01 m/s
b) a = 5.6 m/s²
Explanation:
a)
If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:[tex]\tau = I * \alpha (1)[/tex]
Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:τ = F*r (2)For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).Replacing (2) and (3) in (1), we can solve for α, as follows:[tex]\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)[/tex]
Since the angular acceleration is constant, we can use the following kinematic equation:[tex]\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)[/tex]
Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:[tex]0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)[/tex]
Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:[tex]\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)[/tex]
Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:[tex]v = \omega * r (8)[/tex]
where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.Replacing this value and (7) in (8), we get:[tex]v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)[/tex]
b)
There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:[tex]a_{t} = \alpha * r (9)[/tex]
where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.Replacing this value and (4), in (9), we get:[tex]a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)[/tex]
Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.The radial acceleration is just the centripetal acceleration, that can be expressed as follows:[tex]a_{c} = \omega^{2} * r (11)[/tex]
Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.Replacing this value and (7) in (11) we get:[tex]a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)[/tex]
The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:[tex]a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)[/tex]
When measuring wellness you must consider
A all components of health
B your physical fitness being in the top 10% of the population
C being free of diseases
D both physical and mental health
Answer:
The answer is A) all components of health
Explanation:
Got it right on edge
5.
What is the apparent colour of a red shirt when viewed in pure green light.?
Red
(b)- Green
Yellow (d) Black) (e) Blue
Answer: black
Explanation: When green light is shone on a red object, it absorbs all of the green light and not reflecting anything. Hence, it appears black.
Solve each of the following problems to 3 sig figs and correct Sl units, showing all work.
1. A cart with a mass of 45.0 kg is being pulled to the right with a force of 250 N giving it an
acceleration of 1.30 m/s2. The wheels of the cart are locked and the cart must be dragged.
a) Draw a free body diagram of the cart.
b) Calculate the net force acting on the cart.
c) Create a force table and fill it in.
d) Find the coefficient of kinetic friction.
Answer:
every number to 3 sf = 1) 45.0 2) 250 3) 1.30
Explanation:
your welcome :)
The chart below summarizes the forces applied to four different objects.
Which object will experience the greatest acceleration?
A. Z
B. X
C. Y
D. W
Answer:
C. Y
Explanation:
From Newton's second law of motion, we know that:
Force = mass x acceleration
So;
acceleration = [tex]\frac{Force }{mass}[/tex]
Therefore, to have the highest acceleration at a constant force, the mass must be low. Acceleration is inversely proportional to mass.
Y has the least mass and it will have the highest acceleration
_____________________health is how you feel and how you react to situations bases on how you feel.
Group of answer choices
Mental
Physical
Emotional
Social
Answer: Emotional
Explanation:
Answer:
Physical
Explanation:
I love going going for a jog or a run so does my dog
A rectangular reflecting pool is 85.0 ft wide and 120 ft long. What is the area of the pool in square meters?
In March 1999 the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data back to Earth. Assume a circular orbit with a period of 7.08 × 103 s and orbital speed of 3.40 × 103 m/s . The mass of the GS is 930 kg and the radius of Mars is 3.43 × 106 m. Calculate the mass of Mars.
Answer: [tex]5.944\times 10^{23}\ kg[/tex]
Explanation:
Given
Time period [tex]T=7.08\times 10^3\ s[/tex]
Orbital speed [tex]v=3.40\times 10^3\ m/s[/tex]
mass of GS [tex]m_{GS}=930\ kg[/tex]
Radius of Mars [tex]r=3.43\times 10^6\ m[/tex]
Consider the mass of mars is M
Here, Gravitational pull will provide the centripetal force
[tex]F_G=F_c[/tex]
[tex]\dfrac{GMm_{GS}}{r^2}=\dfrac{m_{GS}v^2}{r}\\M=\dfrac{v^2\cdot r}{G}\\M=\dfrac{(3.43\times 10^3)^2\cdot 3.43\times 10^6}{6.67\times 10^{-11}}[/tex]
[tex]M=5.944\times 10^{23}\ kg[/tex]
In March 1999 the Mars Global Surveyor (GS) entered its final orbit on Mars, sending data back to Earth. The mass of Mars is approximately 6.419 × 10²³ kg.
Kepler's Third Law states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (a) of the orbit:
T² = (4π² / GM) × a³
In a circular orbit, the semi-major axis is equal to the radius of the orbit (r).
Given:
Orbital period (T) = 7.08 × 10³ s
Orbital speed (v) = 3.40 × 10³ m/s
Mass of GS (m) = 930 kg
Radius of Mars (r) = 3.43 × 10⁶ m
The orbital speed (v) is related to the radius (r) and the gravitational constant (G) by:
v = √(GM / r)
v² = GM / r
G = (v² × r) / M
T² = (4π² / [(v² × r) / M]) × r³
T² = (4π² × M × r²) / v²
M = (T² × v²) / (4π² × r²)
M = ( (7.08 × 10³)² × (3.40 × 10³)² ) / (4π² × (3.43 × 10⁶)²)
M = 6.419 × 10²³ kg
Therefore, the mass of Mars is approximately 6.419 × 10²³ kg.
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Let A be the second to last digit and let B be the last two digits of your 8-digit student ID. Example: for 20245347, A = 4 and B = 47.A ball rolls off a table. The table top is 1.2 m above the floor and the ball lands 3.6 m from the base of the table. Determine the speed of the ball at the time it rolled over the edge of the table? Calculate the answer in m/s and rounded to three significant figures.
Answer:
7.35 m/s
Explanation:
Using y - y' = ut - 1/2gt², we find the time it takes the ball to fall from the 1.2 m table top and hit the floor.
y' = initial position of ball = 1.2 m, y = final position of ball = 0 m, u = initial vertical velocity of ball = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for ball to hit the ground.
So, substituting the values of the variables into the equation, we have
y - y' = ut - 1/2gt²
0 - 1.2 m = (0 m/s)t - 1/2(9.8 m/s²)t²
- 1.2 m = 0 - (4.9 m/s²)t²
- 1.2 m = - (4.9 m/s²)t²
t² = - 1.2 m/- (4.9 m/s²)
t² = 0.245 s²
t = √(0.245 s²)
t = 0.49 s
Since d = vt where d = horizontal distance ball moves = 3.6 m, v = horizontal velocity of ball = unknown and t = time it takes ball to land = 0.49 s.
So, d = vt
v = d/t
= 3.6 m/0.49 s
= 7.35 m/s
Since the initial velocity of the ball is 7.35 m/s since the initial vertical velocity is 0 m/s.
It is shown thus V = √(u² + v²)
= √(0² + v²)
= √(0 + v²)
= √v²
= v
= 7.35 m/s
Megan walks 1100\,\text m1100m1100, start text, m, end text to the left in 330\,\text s330s330, start text, s, end text. What was her average speed in \dfrac{\text m}{\text s} s m start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?
Answer:
v = 3.34 m/s
Explanation:
Given that,
Distance, d = 1100 m
Time, t = 330 s
We need to find the average speed of the Megan. It is equal to the total distance divided by total time taken.
[tex]v=\dfrac{1100\ m}{330\ s}\\\\v=3.34\ m/s[/tex]
So, the average speed of Megan is 3.34 m/s.
Answer:
33.3
Explanation:
Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals are made, how much money, to the nearest dollar, would be in the account after 6 years?
Answer:
15448
Explanation:
A=11000(1.01425)^{24}
A=11000(1.01425)
24
Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals are made, the money to the nearest dollar, would be in the account after 6 years is 15448.
What is Compound interest?The compound interest occurs when the interest is reinvested rather than paying it out. It's basically earning interest over interest.
The formula is:
Compound interest, [tex]A = P ( 1 +\frac{r}{n} )^{nt}[/tex]
Where:
A = final Amount
P = initial principal balance
r = interest rate
n = number of times interest applied per time period
t = number of time periods elapsed
Austin invested P=$11000 in an account with an interest rate of r=5.7% = 0.057 (decimal) during t=6 years compounded quarterly. Since there are 4 quarters in a year, n=4.
Thus, Substituting all the values in the given formula,
A = 11000 ( 1 + [tex]\frac{0.057}{4} )^{6*4}[/tex]
= 11000 × 1.4043662796
= 15448.0290
The money to the nearest dollar, would be in the account after 6 years is 15448.
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A block of mass m = 4.4 kg slides from left to right across a frictionless surface with a speed 9.2 m/s It collides in a perfectly elastic collision with a second block of mass M that is at rest. After the collision, the 4.4-kg block reverses direction, and its new speed is 2.5 m/s The block of mass M travels to the right at a speed V of 6.7 m/s what is M
Answer:
[tex]m_2=6.3\:\mathrm{kg}[/tex]
Explanation:
In a perfectly elastic collision, the total kinetic energy of the system is maintained. Therefore, we can set up the following equation:
[tex]\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2[/tex]
Since the second block was initially at rest, [tex]\frac{1}{2}m_2{v_2}^2=0[/tex].
Plugging in all given values, we have:
[tex]\frac{1}{2}m_1{v_1}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2,\\\\\frac{1}{2}\cdot4.4\cdot9.2^2=\frac{1}{2}\cdot 4.4 \cdot (-2.5)^2+\frac{1}{2}\cdot m_2\cdot 6.7^2,\\m_2=\fbox{$6.3\:\mathrm{kg}$}[/tex].
A beaker with water resting on a scale weighs 40 N. A block
suspended on a hanging spring weighs 20 N. The spring scale
reads 15 N when a block is fully submerged in the water. What is
the reading of a scale on which the beaker with water rests, while
the block is submerged in the water after detached from the
hanging spring?
A. 25 N B. 60 N C. 55 N D. 45 N
Answer:
D. 45 N
Explanation:
The weight of the block is 20 N, when the block is fully immerged in water, it weighs 15 N. Hence the loss of weight = 20 N - 15 N = 5 N.
The loss of weight is as a result of the buoyant force. The buoyant force is the upward force exerted by a fluid when an object is fully or partially immersed in a fluid.
The buoyant force of 5 N acts in the upward direction, the weight of the beaker that would be read by the scale when the beaker is immersed in water = 40 N + 5 N = 45 N
In designing buildings to be erected in an area prone to earthquakes, what relationship should the designer try to achieve between the natural frequency of the building and the typical earthquake frequencies?
A) The natural frequency of the building should be exactly the same as typical earthquake frequencies.
B) The natural frequency of the building should be almost the same as typical earthquake frequencies but slightly lower
C) The natural frequency of the building should be very different frem typical earthquake frequencies
D) The natural frequency of the building should be almost the same as typical earthquake frequencies but slightly higher.
Answer:
C) The natural frequency of the building should be very different from typical earthquake frequencies
Explanation:
We shall apply the concept of resonance in this problem .
When a body is applied an external harmonic force ( forced vibration) such that natural frequency of body is equal to frequency of external force or periodicity of external force , the body vibrates under resonance ie its amplitude of vibration becomes very high .
In the present case if natural frequency of building becomes equal to the earthquake's frequency ( external force ) , the building will start vibrating with maximum amplitude , resulting into quick collapse of the whole building . So to avoid this situation , natural frequency of building should be very different from typical earthquake frequencies .
a 1 mole of an ideal gas is kept at 0°C during expansion from 30l to 10l .How much work is done on the gas during expansion
Answer:
20 J
Explanation:
Work done is given force by distance .
W= F * d where F is force given by the product of pressure and area
W= P* Δv where Δv is change in volume.
Given that ;
1 mole of an ideal gas is kept at 0°C, the pressure of the gas is : 1 atm.
Δv is change in volume , 30 l - 10l = 20 l
W= 1 * 20 = 20 J
Two balls of unequal mass are hung from two springs that are not identical. The springs stretch the same distance as the two systems reach equilibrium. Then both springs are compressed and released. Which one oscillates faster?
a) The spring with the heavier ball,
b) Springs oscillate with the same frequency,
c) The spring with the light ball.
Answer:
b) Springs oscillate with the same frequency,
Explanation:
expression for frequency of vibration of mass hanging from a spring is given as follows
f = [tex]\frac{1}{2\pi} \times \sqrt{\frac{k}{m} }[/tex]
k is force constant of spring and m is mass vibrating .
In the present case, if mass stretches the spring by x and remains balanced
mg = k x
[tex]\frac{k}{m} =\frac{g}{x}[/tex]
g and x are same for both cases
[tex]\frac{k}{m}[/tex] will also be same for both cases .
Hence frequency of vibration will also be same for both the balls .
A basketball is picked up off the ground and carried to the top of a platform that is 160 feet up. It is then dropped to the ground. The ball rebounds one-half the height each time it hits the ground. What is the total vertical distance the ball will travel from the moment it is picked up to the moment it reaches its maximum height after the fourth bounce
Answer:
Explanation:
The total distance travelled by ball before first bounce
= 160 + 160 = 320 ft
Distance travelled between first bounce and second bounce
= 80 + 80 = 160 ft
Distance travelled between second bounce and third bounce
= 40 + 40 = 80 ft
Distance travelled between third bounce and fourth bounce
= 20 + 20 = 40 ft .
Distance travelled in fourth bounce = 10 ft
Total distance travelled = 320 + 160 + 80 + 40 + 10
= 610 ft .
A ray of monochromatic light is incident on a plane mirror at and angle of 30. The angle of reflection for the light is
1)15
2)30
3)60
4)90
Answer:
30 degrees
Explanation: reflection, same angle
For a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
Reflection occurs when radiation bounces off from a surface. Light is an electromagnetic wave and it can be reflected. According to the laws of reflection, the angle of incidence is equal to the law of reflection.
Hence, for a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
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Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the Earth.
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Explanation:
Where do the rain forest plants get their nutrients?
from plants that fall to the forest floor
from animals that die on the forest floor
from plants and animals that die and fall to forest floor
from sunlight
Answer:
From plants and animals that die and fall to forests floor
Explanation:
Hope it helps
Mark brainliest
true or false A person's speed around the Earth is faster at the poles than it is at the equator.
Answer:False
Explanation:The Earth rotates faster at the equator than at the poles.
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.69 times a second. A tack is stuck in the tire at a distance of 0.331 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed.
Answer:
the tack's tangential speed is 5.59 m/s
Explanation:
Given that;
R = 0.331 m
wheel rotates 2.69 times a second which means, the wheel complete 2.69 revolutions in a second, so
ω = 2.69 rev/s × 2π/1s = 16.9 rad/s
using the relation of angular speed with tangential speed
tangential speed v of the tack is expressed as;
v = R × ω
so we substitute
v = 0.331 m × 16.9 rad/s
v = 5.59 m/s
Therefore, the tack's tangential speed is 5.59 m/s