Suppose that a 710 kg car is traveling at 13 m/s. Its brakes can apply a force of 4000 N.

Consult Multiple-Concept Example 5 for insight into solving this problem. A skier slides horizontally along the snow for a dista

antoniya [11.8K]

The initial velocity of the skier is 4.5 m/s

Explanation:

There is only one force acting on the skier in the horizontal direction, and it is the force of friction, whose magnitude is

$F_f = -\mu_k mg$

where

$\mu_k = 0.050$ is the coefficient of kinetic friction

m is the mass of the skier

$g=9.8 m/s^2$ is the acceleration of gravity

The negative sign is due to the fact that the direction of the force of friction is opposite to the direction of motion of the skier.

According to Newton's second law, the net force acting on the skier is equal to the product between his mass and his acceleration, so we can write:

$F=ma\\ \rightarrow -\mu_k mg = ma$

So, the acceleration of the skier is

$a=-\mu_k g$

Now we can apply the following suvat equation to find the initial velocity of the skier:

$v^2-u^2=2as$

where

v = 0 is the final velocity (he comes to a stop)

u is the initial velocity

s = 21 m is the displacement

a is the acceleration

Substituting the equation for the acceleration and solving for u, we find

$u=\sqrt{v^2-2as}=\sqrt{-2(-\mu_k g) s}=\sqrt{2(0.050)(9.8)(21)}=4.5 m/s$

Learn more about accelerated motion here:

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5 0

3 years ago

special metal hangers or stirrips called joist hangers are used when joist must be _____ the bottom of the grider or beam

djyliett [7]

Answer:

Special metal hangers or stirrups called joist hangers are used when joist must be flush with the bottom of the girder or beam.

Explanation:

A joist hanger also known as a beam hanger is a mechanical device which is used to fasten joists and rafters.
The rafters are the carried members to beams and headers are the carrying members.

Thus, special metal hangers or stirrups called joist hangers are used when joist must be flush with the bottom of the girder or beam.

Learn more about construction here:

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7 0

1 year ago

Calculate the gravitational force between a 10 kg ball and 20 kg ball

elena-s [515]

Answer:

$5.34\cdot 10^{-10} N$

Explanation:

The gravitational force is an attractive force exerted between any objects with mass.

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the two objects

In this problem, we have:

$m_1 = 10 kg$ is the mass of the 1st ball

$m_2 = 20 kg$ is the mass of the 2nd ball

r = 5 m is the separation

So, the gravitational force between them is

$F=(6.67\cdot 10^{-11})\frac{(10)(20)}{5^2}=5.34\cdot 10^{-10} N$

7 0

3 years ago

A train is traveling at 30.0 m/sm/s relative to the ground in still air. The frequency of the note emitted by the train whistle

Zigmanuir [339]

Answer

given,

speed of sound = 344 m/s

speed of train = 30 m/s

frequency emitted by the train = 262 Hz

Doppler's effect

$f_L = \dfrac{v + v_L}{v + v_s}\ f_S$

f_L is the frequency of listener

f_S is the frequency of the source of the sound

v is the speed of the sound

v_L is the speed of listener.

v_S is the speed of the source

a) Speed of the passenger in another train , v = 18 m/s

another train is moving in opposite direction and approaching

v_L is positive as the listener is moving forward.

v_S is negative at the source is moving toward the listener.

$f_L = \dfrac{344 + 18}{344 - 30}\times 262$

$f_L = 302\ Hz$

b) Speed of the passenger in another train , v = 18 m/s

another train is moving in opposite direction and receding

v_L is negative as the listener is moving away from source.

v_S is positive at the source is moving away the listener.

$f_L = \dfrac{344 - 18}{344 + 30}\times 262$

$f_L = 228.37\ Hz$

8 0

3 years ago

A civil engineer plans to design a curved ramp such that a car may not have to rely on friction to round the curve without skidd

Delicious77 [7]

Answer: $\theta =\tan ^{-1}(\frac{v^2}{gR})$

Explanation:

let \theta be the inclination at which curve is tilted

v is the speed of car and R is the radius of curve

m is the mass of car

Suppose R is the reaction offered by road to car

Resolving R in x and y direction we get

$R\cos \theta$ will balance weight and $R\sin \theta$ will provide the necessary centripetal Force

thus $R\sin \theta =\frac{mv^2}{R}$ ------------1

$R\cos \theta =m g$ ----------------2

Divide 1 & 2 we get

$\frac{R\sin \theta }{R\cos \theta }=\frac{mv^2}{mgR}$

$\tan \theta =\frac{v^2}{gR}$

$\theta =\tan ^{-1}(\frac{v^2}{gR})$

8 0

3 years ago