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Alex777 [14]
3 years ago
10

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force

of static friction f on the block?
A) f > mg
B) f = mgsin(?)
C) f > mgcos(?)
D) f = mgcos(?)
E) f > mgsin(?)

Physics
1 answer:
nignag [31]3 years ago
8 0

Answer:

Option B

Explanation:

For a system of block on inclined ramp shown in the attached image. From the attached image, the Normal force N, weight mg and frictional force f act on the block.  The sum of vertical forces should be zero just as sum of vertical forces should be zero when the system is in equilibrium condition.

Taking sum of forces along the inclined plane we deduce that  

[tex]f=mgsin \theta [tex]

Therefore, option B is the correct option.

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A 92kg astronaut and a 1200kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving
Harman [31]

Answer:

13.7m

Explanation:

Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.

After the push

m_av_a + m_sv_s = 0

Where m_a = 92kg is the mass of the astronaut, m_s = 1200kg is the mass of the satellite, v_s = 0.14 m/s is the speed of the satellite. We can calculate the speed v_a of the astronaut:

v_a = \frac{-m_sv_s}{m_a} = \frac{-1200*0.14}{92} = -1.83 m/s

So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be

d = vt = 1.83 * 7.5 = 13.7 m

4 0
3 years ago
In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t
galina1969 [7]

Answer:

W_x = 0.9156\ c

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}

u_x = 0.543 c

v_x = - 0.741 c

W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}

W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}

W_x = \dfrac{1.284c}{1+0.402363}

W_x = 0.9156\ c

Relative velocity of the particle is W_x = 0.9156\ c

5 0
3 years ago
A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,
fomenos

Answer:

The momentum would be doubled

Explanation:

The magnitude of the momentum of the freight train is given by:

p=mv

where

m is the mass of the train

v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

m'=2m

therefore, the new momentum is

p'=m'v=(2m)v=2(mv)=2p

so, the momentum has also doubled.

7 0
3 years ago
an elevator mass of 7700 kg falls from a height of 32 m after a sudden failure in the hoisting cable. The mass is stopped by a s
valkas [14]

Answer:k=28.29 kN/m

Explanation:

Given

mass m =7700 kg

height from which Elevator falls h=32 m

Let x be the compression in the spring

thus From conservation of Energy Potential energy will convert in to Elastic Potential Energy of spring

\frac{kx^2}{2}=mg(h+x)----------1

also maximum acceleration is 5g

thus

mg-kx=ma

here a=-5g

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Substitute x in equation 1

0.5\times k\times (\frac{6mg}{k})^2=mg(h+\frac{6mg}{k})

18\frac{(mg)^2}{k}=mgh+6\frac{(mg)^2}{k}

k=12\cdot \frac{mg}{h}

k=12\times \frac{7700\times 9.8}{32}

k=28.29 kN/m

4 0
3 years ago
Describe how the student should measure the time taken for the toy parachute to
PolarNik [594]

The time must be measured with respect to gravity. As it falls, it has free fall that is the force acting on it will be the gravity.With the distance in account, d = 1/2 gt²

t = √(2d/g)

4 0
3 years ago
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