You are pulling a friend on a sled with a constant velocity using a rope held
horizontally in the snow. Using a spring scale attached to the rope, you
measure the tension to be 112 N. The combined mass of the sled and your
friend is 57.4 kg. Write the surface force the ground exerts on the sled in
component form, and determine its magnitude.

The velocity of the tennis ball when it hits the ground is calculated as 4.43m/s. Details about velocity can be found below.

The velocity of a moving object can be calculated using the equation of motion as follows:

v² = u² + 2as

Where;

v² = 0² + 2(9.8) (1)

v² = 19.6

v = ✓19.6

v = 4.43m/s

Therefore, the velocity of the tennis ball when it hits the ground is calculated as 4.43m/s.

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Answer:

d. distance travelled

Explanation:

answer

[tex]\\ \sf\longmapsto m1v1+m2v2=(m1+m2)v3[/tex]

[tex]\\ \sf\longmapsto 7(17)+8(-14)=(7+8)v3[/tex]

[tex]\\ \sf\longmapsto 119-112=15v3[/tex]

[tex]\\ \sf\longmapsto 15v3=7[/tex]

[tex]\\ \sf\longmapsto v3=7/15[/tex]

[tex]\\ \sf\longmapsto v3=2.1m/s[/tex]

Hi there!

To find the appropriate force needed to keep the block moving at a constant speed, we must use the dynamic friction force since the block would be in motion.

Recall:

[tex]\large\boxed{F_D = \mu N}}[/tex]

The normal force of an object on an inclined plane is equivalent to the vertical component of its weight vector. However, the horizontal force applied contains a vertical component that contributes to this normal force.

[tex]\large\boxed{N = Mgcos\theta + Fsin\theta}}[/tex]

We can plug in the known values to solve for one part of the normal force:

N = (1)(9.8)(cos30)  + F(.5) = 8.49  + .5F

Now, we can plug this into the equation for the dynamic friction force:

Fd= (0.2)(8.49 + .5F) = 1.697 N + .1F

For a block to move with constant speed, the summation of forces must be equivalent to 0 N.

If a HORIZONTAL force is applied to the block, its horizontal component must be EQUIVALENT to the friction force. (∑F = 0 N). Thus:

Fcosθ = 1.697 + .1F

Solve for F:

Fcos(30) - .1F = 1.697

F(cos(30) - .1) = 1.697

F = 2.216 N

Answer:

Usually, a solution can have several criteria and constraints. Even though all are important, some criteria are more important than others. The same holds true for constraints. But what do you do if it's impossible for a solution to cover every criterion while avoiding every constraint? In cases like this, you can use prioritization. Listing criteria and constraints based on priority shows the relative importance of each. You will need to prioritize the criteria and constraints for each sub-problem so that you can design a solution for each one individually. Prioritization can help you compare two different possible solutions. For example, the criterion that cars travel at 15 mph through the neighborhood might be a higher priority than the constraint that homeowners are only willing to spend $10,000 on this issue. If this is the case, you would want to generate solutions that also follow the priority in mind. All criteria are important, but engineers must sometimes make a trade-off, which is a compromise or change in one or more criteria or constraints so that they can be met at the same time. This is where prioritization comes in handy as it helps determine the trade-offs. A solution that is doing a better job of meeting one criterion may result in not completely meeting another criterion. Prioritization will help you choose which solution to go with.

Explanation:

I got this from quizlet :)

Answer:it was lifted

Explanation:

by me lifting it because i can life it very high

Answer:

The force of gravity on earth, no matter the object is approximately 9.8 m/ s2 . The reason the crumpled paper hits the ground first is because of air resistance. A crumpled piece of paper has less surface area than an piece of paper that is not crumpled. More surface area means more air resistance.

The velocity of the pitcher at the given mass is 0.1 m/s.

The given parameters:

Let the velocity of the pitcher = u₁

Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;

m₁u₁ = m₂u₂

[tex]u_1 = \frac{m_2 u_2}{m_1} \\\\u_1 = \frac{0.15 \times 35}{50} \\\\u_1 = 0.105 \ m/s\\\\u_1 \approx 0.1 \ m/s[/tex]

Thus, the velocity of the pitcher at the given mass is 0.1 m/s.

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Answer:

A-Caclcuate the potential energy of the ball at that height

Explanation:

(a). Mass of the Body = 10 kg.

Height = 10 m.

Acceleration due to gravity = 9.8 m/s².

Using the Formula,Potential Energy = mgh

= 10 × 9.8 × 10 = 980 J.

(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.

∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.

∴ Kinetic Energy = 980 J.

(c). Kinetic Energy = 980 J.

Mass of the ball = 10 kg.

∵ K.E. = 1/2 × mv²

∴ 980 = 1/2 × 10 × v²

∴ v² = 980/5

⇒ v² = 196

∴ v = 14 m/s.

Answer:

the mass of the body is 2.75 kg

Gravitational Force

Explanation:-

The magma flows through the gap between the rocks which are stacked up on top of one another.

The gravity attracts the magma to do that.

Also the rocks are stacked because of the gravitational force between them

Hence Option D is correct

Hope it helps

Answer:

pls mark me as brainliest

Explanation:

last one

The initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.

CALCULATE INITIAL VELOCITY:

The initial velocity of the car can be calculated by using one of the equation of motion as follows:

V = u + at

Where;

According to this question, a car accelerates at a constant rate of 3 m/s² for 5 seconds. If it reaches a velocity of 27 m/s, its initial velocity is calculated as follows:

u = v - at

u = 27 - 3(5)

u = 27 - 15

u = 12m/s.

Therefore, the initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.

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Answer:

actually you can't find it without acceleration.

you need to use one of these three equations

1- Vf = Vi + at

2- 2ad = Vf^2 - Vi^2

3- d = Vit + 1/2 at^2

Applying Charles law

[tex]\\ \sf\Rrightarrow \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}[/tex]

[tex]\\ \sf\Rrightarrow \dfrac{100}{313}=\dfrac{V_2}{263}[/tex]

[tex]\\ \sf\Rrightarrow V_2=\dfrac{26300}{313}[/tex]

[tex]\\ \sf\Rrightarrow V_2=84.02ml[/tex]

Answer:

[tex]0.6\; \rm m\cdot s^{-1}[/tex], assuming that drag of the water on the log is negligible.

Explanation:

The momentum [tex]p[/tex] of an object is equal to the product of mass [tex]m[/tex] and velocity [tex]v[/tex]. That is, [tex]p = m\, v[/tex].

If the drag of water on the log is negligible, momentum of the beaver and the log, combined, would be preserved.

The momentum of the beaver and the log, combined, was initially [tex]0\; \rm kg\cdot m \cdot s^{-1}[/tex].

The momentum of the beaver right after the dive would be [tex]7.5 \; {\rm kg} \times 4\; {\rm m \cdot s^{-1}} = 30\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

The sum of the momentum of the beaver and the log is conserved and should continue to be [tex]0\; {\rm kg\cdot m \cdot s^{-1}}[/tex] even after the dive. Since the momentum of the beaver is [tex]30\; {\rm kg \cdot m \cdot s^{-1}[/tex] after the dive, the momentum of the log should become [tex](-30)\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Since the mass of the log is [tex]50\; {\rm kg}[/tex], the new velocity of this log would be:

[tex]\begin{aligned}v &= \frac{p}{m} \\ &= \frac{(-30)\; {\rm kg \cdot m \cdot s^{-1}}}{50\; {\rm kg}} \\ &= (-0.6)\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].

(The new velocity of the log is negative because the log would be moving away from the beaver.)

The speed of an object is the magnitude of velocity. For this log, a velocity of [tex](-0.6)\; {\rm m \cdot s^{-1}}[/tex] would correspond to a speed of [tex]|(-0.6)\; {\rm m \cdot s^{-1}| = 0.6\; {\rm m \cdot s^{-1}[/tex].

The major characteristic an organic compound has which an inorganic

compound doesn't  have is the presence of carbon atoms in the molecule.

This helps us to differentiate the compounds into their respective divisions.

Inorganic compounds

Organic compounds

We can observe that inorganic compounds lack carbon atoms which makes

the answer valid.

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Answer:

B and E because yk thats the asnwer

Answer:

TRUE !!

Explanation:

An atom is the smallest particle of an element

Answer: 25N to the right

Explanation: 35N - 10N = 25N

Answer: 0.22

Explanation: Subtract 0.01kg from 0.23kg

Let D be the total distance (say in meters) traveled by the train and T the time (say in seconds) it takes to do so. (Assume the train moves in a straight line in only one direction.) Then the average velocity of the train as it covers this distance is

v (ave) = D/T

We're told the train can traverse a distance of D/4 in a matter of T/2 seconds if it moves at a speed of 5 m/s. This means

D/4 = (5 m/s) (T/2)

⇒   5 m/s = 1/2 D/T

⇒   v (ave) = D/T = 10 m/s

Answer:

C or 3

Explanation:

A: no they are called sources of sound. A is incorrect.

B: It does. Many people attest to this. But this is not a property of physics.

C: A media is required is the correct answer.

D: Dogs might. In general we don't.

I would pick C

Answer:

1,967 kilojoules.

Explanation:

1 kilo = a thousand.

There are 1.967 kilojoules in 1,967,256 joules.

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Answer:

wat

Explanation:

Gravitational force is responsible for how magma flows and rocks stack up on top of one another.

Hence, it is the gravitational force that is responsible for how magma flows and rocks stack up on top of one another.

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Answer:

Metals conduct heat and reduce the kinetic energy within the components that need to remain cool

Range be R and height be h

[tex]\boxed{\sf R=\dfrac{u^2sin2\theta}{g}}[/tex]

[tex]\boxed{\sf h=\dfrac{u^2sin^2\theta}{2g}}[/tex]

u=initial velocity

theta is angle of projection.

g=acceleration due to gravity

ATQ

[tex]\\ \sf\longmapsto R=2h[/tex]

[tex]\\ \sf\longmapsto \dfrac{u^2sin2\theta}{g}=\dfrac{2u^2sin^2\theta}{2g}[/tex]

Cancelling required ones

[tex]\\ \sf\longmapsto sin^2\theta=sin2\theta[/tex]

sin2O=2sinOcosO

[tex]\\ \sf\longmapsto sin^2\theta=2sin\theta cos\theta [/tex]

[tex]\\ \sf\longmapsto \dfrac{sin^2\theta}{sin\theta cos\theta=2[/tex]

[tex]\\ \sf\longmapsto \dfrac{sin\theta}{cos\theta}=2[/tex]

[tex]\\ \sf\longmapsto tan\theta=2[/tex]

[tex]\\ \sf\longmapsto \theta=tan^{-1}(2)[/tex]

[tex]\\ \sf\longmapsto \theta=63.4°[/tex]

Done

Option B is correct