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3 years ago

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A block is being pulled upward along an inclined surface at a constant speed. Which of the following statements is correct? Grou

p of answer choices The net force on the block points upward along the inclined surface. The net force on the block is zero. The net force on the block is larger than the magnitude of the weight of the block. The net force on the block is smaller than the magnitude of the weight of the block but greater than zero.

1 answer:

Alex17521 [72]3 years ago

5 0

Answer:The net force on the block is zero.

Explanation:

Given

Block is being pulled upward along an inclined surface at a constant speed

As speed is constant and moved in a straight line along the plane therefore its velocity is also constant .

and change in velocity is equal to acceleration therefore acceleration is zero here i.e. net force is zero acting on the body.

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How do you draw an ear

Neporo4naja [7]

Answer:

drawing of ear step by step

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3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff

postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

$F_c=\frac{mv^2}{R}$

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

$N=mg$

Therefore, frictional force, $f=\mu mg$

Now, frictional force = centripetal force

$f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}$

Plug in the given values and solve for 'μ'. This gives,

$\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833$

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0

4 years ago

A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m

lbvjy [14]

The equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.

The horizontal momentum is given by:

$p_{i} = p_{f}$

$m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}$

Where:

m₁: is the mass of the lab cart = 15 kg

m₂: is the <em>mass </em>of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the<em> lab cart </em>

$v_{2}_{i}$ : is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the<em> lab cart </em>

$v_{2}_{f}$ : is the <em>final velocity</em> of the <em>object </em>

Then, the horizontal momentum is:

$15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}$

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

$15v_{1}_{i} + 2*0 = v_{f}(15 + 2)$

Therefore, the equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ represents the horizontal momentum (option 3).

Learn more about linear momentum here:

brainly.com/question/2141713?referrer=searchResults

brainly.com/question/2400186?referrer=searchResults

I hope it helps you!

4 0

3 years ago

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

$m_1$ = Mass of first car = 120 kg

$m_2$ = Mass of second car

$u_1$ = Initial Velocity of first car = 14 m/s

$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0

4 years ago

A classmate is attempting to draw a free body diagram for a box being pushed across the floor at a constant speed. The image bel

aivan3 [116]

Answer:

A. The applied force should be the same size as the friction force

Explanation:

Whenever we apply a force to an object it moves if the force applied to that object is unbalanced and there is no force or a lesser force to counter it. According to Newton's Second Law of motion, when an unbalanced force is applied to an object it produces an acceleration in the object in its own direction. So, the two forces acting on this box are the frictional force and the applied force in horizontal direction. In order to move the box at constant speed, the applied force must first, overcome the frictional force, so the object can start its motion. Since, the motion has constant velocity, it means no acceleration. So, the force must be balanced in order to avoid acceleration as a consequence of Newton's Second Law of motion. Therefore, the correction in this case will be:

<u>A. The applied force should be the same size as the friction force</u>

7 0

3 years ago