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yulyashka [42]
3 years ago
10

YALL HELP ME QUICK IN 10 MINS WITH WORKING OUT

Physics
2 answers:
svetlana [45]3 years ago
8 0

Answer:

bro I'm sorry I'm only in 7th grade

Romashka-Z-Leto [24]3 years ago
4 0
The top is kinda cut off but it says “weight =mg

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Example 4:
earnstyle [38]

Answer:

Given speed of light in diamond =1.24*10m/s

speed of light (c) =3*10^8m/s

refractive index of diamond =3*10^8/1.24*10=2.41*10^7m/s

3 0
3 years ago
Friction removes energy from objects in motion. Which statement best describes how this works?
xenn [34]

<em>friction transforms KE into thermal energy (a)</em>

That's why, if it goes on long enough, the moving object actually gets warm.

8 0
3 years ago
Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From the frame of reference of car 1, what is the veloci
EleoNora [17]
We subtract the velocity of car 1 from the velocity of car 2:
v=(30\ mph\ North)-(20\ mph\ South)
=(30\ mph\ North)+(20\ mph\ North)
=50\ mph\ North
4 0
3 years ago
Read 2 more answers
When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some poi
eimsori [14]

Answer: The force of kinetic friction is smaller than that of static friction, but F_g  remains the same.

Explanation:

The situation is same as when a book is pushed with an increasing force on a table; When the force is low, book doesn't move, until that under a given force starts moving, and then it goes on movement even if the force decreases a bit.

The physical explanation for this, that friction force adopts any value needed to avoid to move the object, till a limit value is achieved, called static friction force, equal to the normal force times the static friction coefficient.

Once in movement, the kinetic friction coefficient replaces the static one , and  in general is lower than the static one, so the force diminishes.

In the case of the box sliding down the board, the force that tries to move the object down the board, is the component of the weight parallel to the board, that can be showed that being equal to the weight times the sinus of the angle of the board with the horizontal, as follows:

F_g = m g sin θ

When θ increases, F_g does the same, so friction force always has the same magnitude than F_g (but opposite direction) so the box doesn't move, till that θ takes a value that produces a F_g equal to static friction force.

Beyond this limit angle, F_g (remaining the same for a given angle) is greater than the kinetic friction force, and the box slides.

In the limit, when θ=90º, sin θ =1⇒ F_g = mg, so the object is in free fall.

6 0
3 years ago
A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×
HACTEHA [7]

Answer:

f_r = 150.47 N

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta

∑ F_y = 0

0 = N cos \theta - f_r sin \theta - mg

N = \dfrac{f_rsin \theta + mg}{cos \theta}

\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta

              = f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta

        f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}

         f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}

f_r = 150.47 N

8 0
3 years ago
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