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Zepler [3.9K]
4 years ago
8

A 2.0-N force acts horizontally on a 10-N block that is initially at rest on a horizontal surface. The coefficient of static fri

ction between the block and the surface is 0.50. Suppose that the block now moves across the surface with constant speed under the action of a horizontal 3.0-N force. Which one of the following statements concerning this situation is false? The net force on the block is zero newtons. The coefficient of kinetic friction between the block and the surface is 0.30. The frictional force on the block has magnitude 3.0 N. The direction of the total force that the surface exerts on the block is vertically upward. The block is not accelerated.
Physics
1 answer:
Nady [450]4 years ago
7 0

Answer:

None of the statement is false - all of them are true

Explanation:

Let's analyze each statement:

- The net force on the block is zero newtons. --> TRUE. In fact, the block is moving across the surface at constant speed: constant speed means zero acceleration, and according to Newton's second law,

F = ma

this also means a net force of zero newtons.

- The frictional force on the block has magnitude 3.0 N. --> TRUE. The horizontal acceleration of the block is zero, so the resultant of the horizontal forces must be zero:

F-F_f = 0

where F = 3.0 N is the horizontal push and F_f is the frictional force. From the equation, we find

F_f = F = 3.0 N

- The coefficient of kinetic friction between the block and the surface is 0.30. --> TRUE. The frictional force is 3.0 N, and its expression is

F_f = \mu_k (mg)

where \mu_k is the coefficient of kinetic friction and (mg)=10 N is the weight of the block. Solving for \mu_k, we find

\mu_k = \frac{F_f}{mg}=\frac{3 N}{10 N}=0.30

- The direction of the total force that the surface exerts on the block is vertically upward. --> TRUE. Since gravity is acting downward, and the block is not accelerating on the vertical direction neither, there must be an equal and opposite force acting upward on the block: and this force is the force exerted by the surface on the block.

- The block is not accelerated. --> TRUE: the block is moving at constant speed, so its acceleration is zero.

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Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.

5 0
2 years ago
A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is
Valentin [98]

Answer:

<em>The end of the ramp is 38.416 m high</em>

Explanation:

<u>Horizontal Motion </u>

When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

\displaystyle d=v\cdot\sqrt{\frac  {2h}{g}}

If the maximum horizontal distance is known, we can solve the above equation for h:

\displaystyle h=\frac  {d^2g}{2v^2}

The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:

\displaystyle h=\frac  {70^2\cdot 9.8}{2\cdot 25^2}

\displaystyle h=\frac  {4900\cdot 9.8}{2\cdot 625}

h= 38.416 m

The end of the ramp is 38.416 m high

8 0
2 years ago
Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i
densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0
3 years ago
A key falls from a bridge that is 32 m above the water. It falls directly into a model boat, moving with constant velocity, that
FinnZ [79.3K]

Answer:

Speed of the boat, v = 4.31 m/s

Explanation:

Given that,

Height of the bridge, h = 32 m

The model boat is 11 m from the point of impact when the key was released, d = 11 m

Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :

s=ut+\dfrac{1}{2}at^2

Here, u = 0 and a = g

t=\sqrt{\dfrac{2s}{g}}

t=\sqrt{\dfrac{2\times 32}{9.8}}

t = 2.55 seconds

Let v is the speed of the boat. It can be calculated as :

v=\dfrac{d}{t}

v=\dfrac{11\ m}{2.55\ s}

v = 4.31 m/s

So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.

3 0
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What is a sedimentary, igneous, and a metamorphic rock?
Tanya [424]

Answer:

Igneous rock , formed by the cooling of magma (molten rock) inside the Earth or on the surface. Sedimentary rocks, formed from the products of weathering by cementation or precipitation on the Earth's surface. Metamorphic rocks, formed by temperature and pressure changes inside the Earth.

Explanation:

The information was found on:

https://msnucleus.org/membership/html/k-6/rc/rocks/3/rcr3_1a.html

6 0
3 years ago
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