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2 years ago

When a student stands on a rotating table,the frictional force exerted on the student by the table is?

Physics

1 answer:

lakkis [162]2 years ago

7 0

Answer:

Less

Explanation:

because static friction is more than rolling friction

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A satellite that orbits Earth with a speed of v0 must be in an orbit of radius 8RE to maintain a circular orbit, where RE is the

NISA [10]

Answer:

1.024 × 10⁸ m

Explanation:

The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.

So, ω = v₀/8RE

For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have

2v₀ = Rω

substituting ω = v₀/8RE into the equation, we have

2v₀ = v₀R/8RE

dividing both sides by v₀, we have

2v₀/v₀ = R/8RE

2 = R/8RE

So, R = 2 × 8RE

R = 16RE

substituting RE = 6.4 × 10⁶ m

R = 16RE

= 16 × 6.4 × 10⁶ m

= 102.4 × 10⁶ m

= 1.024 × 10⁸ m

8 0

2 years ago

Suppose two astronauts on a spacewalk are floating motionless in space, 3.0 m apart. Astronaut B tosses a 15.0 kg IMAX camera to

marta [7]

Answer:

$\frac{ 112.5}{15+m_{A}}=v_{f}$

(we need the mass of the astronaut A)

Explanation:

We can solve this by using the conservation law of the linear momentum P. First we need to represent every mass as a particle. Also we can simplify this system of particles by considering only the astronaut A with an initial speed $v_{iA}$ of 0 m/s and a mass $m_{A}$ and the IMAX camera with an initial speed $v_{ic}$ of 7.5 m/s and a mass $m_{c}$ of 15.0 kg.

The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

$P_{i}=p_{ic}+p_{iA}\\P_{i}=m_{c}v_{ic}+m_{A} v_{iA}\\P_{i}=15*7.5 + m_{A}*0\\P_{i}=112.5 \frac{kg.m}{s}$

By the law of conservation we know that $P_{i} =P_{f}$

For $P_{f}$ (final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).

So:

$P_{i} =P_{f}=112.5\\$

$112.5=(m_{c}+m_{A})v_{f}\\\frac{ 112.5}{m_{c}+m_{A}}=v_{f}\\\frac{ 112.5}{15+m_{A}}=v_{f}$

3 0

2 years ago

Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster

Masja [62]

Answer:

1.10261 times g

416.17506 mph

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g$

The acceleration is 1.10261 times g

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s$

In mph

$186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph$

The speed of the dragster is 416.17506 mph

5 0

2 years ago

An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

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

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$