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3 years ago

How does the direction of friction relate to the direction of motion

1 answer:

5 0

It reacts to the motion. If the mass hanging from the pulley was overwhelmingly heavier than the mass on the ramp, it'll obviously pull the ramp mass up and thus friction would be trying to oppose this and vice versa.

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Astronaut Rob leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. Rob's destination is a star sys

olchik [2.2K]

Answer:

A) 15.0 years

Explanation:

Due to the distance to the star system is in light-year units, we can compute the time by using:

$t=\frac{d}{v}=\frac{14.4 l-y}{0.960}=15l-y$

then, Rob will take to complete the trip about 15 light-years.

hope this helps!!

7 0

3 years ago

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"

Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

4 0

3 years ago

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

Explanation:

Given that,

Mass of proton $m=1.67\times10^{-27}\ kg$

Speed $v= 9.00\times10^{7}\ m/s$

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2$

$K.E=6.76\times10^{-12}\ J$

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

$K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)$

$K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)$

$K.E=7.25\times10^{-12}\ m/s$

Hence, The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

7 0

3 years ago

Anna drives north at a speed of 50 km/h for the first hour. Then, she drives north for a second hour but slows down to 30 km/h.

8090 [49]

Answer:

The correct answer is A The distance is greater in the first hour because her speed is faster.

Explanation:

During the first hour, Anna is driving at a speed of 50 km/h. During the second hour, she is only driving at a speed of 30 km/h. The faster she goes, the farther she will go.

Hope this helps,

♥A.W.E.S.W.A.N.♥

8 0

3 years ago

Read 2 more answers

How is the electrostatic force affected when the magnitude of a charge is doubled?

BaLLatris [955]

The magnitude of the electrostatic force between two charges is given by:
$F=k_e \frac{q_1 q_2}{r^2}$
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) The magnitude of the electrostatic force doubles

4 0

3 years ago