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Zarrin [17]
3 years ago
8

Read each scenario below. Then select the answer choices that complete the sentences,

Physics
2 answers:
Likurg_2 [28]3 years ago
6 0

Answer:

More

Less

Less

More

Explanation:

i got all correct

12345 [234]3 years ago
5 0

Answer:

Higher

Lesser

Lesser

Longer

Explanation:

Search...

54

ryanzl6659

14 hours ago

Physics

College

Read each scenario below. Then select the answer choices that complete the sentences,

A car engine has HIGHER power than a house because a car engine does the same amount of work in LESSER time,

Yaserin and Raj each had 10 tones of equal weight to stack next to each other on the same shelt, at the same height and in the same arrangement. Vasarin completed the task in 2 minutes, wille Raj took 3 minutes to stack his bles,

Raj applied LESSER power than Yaskrin because his stacking took LONGER to do the same amount of

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Electrostatics is described as the study of electric charges that can be collected and held in place. Is this true or false?
Rasek [7]

Electrostatics is described as the study of electric charges that can be collected and held in place.

The above given statement is a true st

7 0
1 year ago
A 1.20 × 104 kg railroad car moving at 7.70 m/s to the north collides with and sticks to another railroad car of the same mass t
saul85 [17]

Answer: 4.77m/s

Explanation:

According to the law of conservation of momentum which states that the sum total of momentum of bodies before collision is equal to the sum of their momentum after collision. Note that the two bodies will move at a common velocity after colliding.

Let m1 and m2 be the mass of the first and second railroad cars

u1 and u2 be the velocities of the railroad cars

v be the common velocity

Using the formula

m1u1 + m2u2 = (m1 +m2)

m1 = 1.20×10⁴kg

m2 = 1.20×10⁴kg (body of same mass)

u1 = 7.70m/s

u2 = 1.84m/s

v = ?

(1.20×10⁴×7.7) + (1.20×10⁴×1.84) = (1.20×10⁴ + 1.20× 10⁴)v

9.24×10⁴ + 2.21×10⁴ = 2.4×10⁴v

11.45×10⁴ = 2.4×10⁴v

v = 11.45×10⁴/2.4×10⁴

v = 4.77m/s

The velocity of the cars after collision will be 4.77m/s

5 0
3 years ago
HELPPPPP ASAP WILL GIVE brainly to correct answer
viva [34]

Answer:

Option d

Plate tectonics i think but im not sure

3 0
3 years ago
Read 2 more answers
. If she
I am Lyosha [343]

Answer:

9 meters

Explanation:

Given:

Mass of Avi is, m=40\ kg

Spring constant is, k=176,400\ N/m

Compression in the spring is, x=20\ cm=0.20\ m

Let the maximum height reached be 'h' m.

Now, as the spring is compressed, there is elastic potential energy stored in the spring. This elastic potential energy is transferred to Avi in the form of gravitational potential energy.

So, by law of conservation of energy, decrease in elastic potential energy is equal to increase in gravitational potential energy.

Decrease in elastic potential energy is given as:

EPE=\frac{1}{2}kx^2\\EPE=\frac{1}{2}\times 176400\times (0.20)^2\\EPE=88200\times 0.04=3528\ J

Now, increase in gravitational potential energy is given as:

GPE=mgh=40\times 9.8\times h=392h

Now, increase in gravitational potential energy is equal decrease in elastic potential energy. Therefore,

392h=3528\\\\h=\frac{3528}{392}\\\\h=9\ m

Therefore, Avi will reach a maximum height of 9 meters.

6 0
4 years ago
Alice's friends Bob and Charlie are having a race to a distant star 10 light years away. Alice is the race official who stays on
hodyreva [135]

Solution :

The distance between the starting point and the end point, L_0 = 10 light years

But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.

For Bob,

Speed of Bob's rocket with respect to Alice, L_b = 0.7  \ c

So the distance appeared to Bob due to the length contraction,

$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$

$L_b=10\times \sqrt{1-0.49} \ Ly$

    $=7.1 \ Ly$

Therefore, the time required to finish the race by Bob is

$t_b = \frac{L_b}{V_b}$

  $=\frac{7.1 \ c}{0.7 \ c}$

  = 10.143 year

For Charlie,

Speed of Charlie's rocket with respect to Alice, L_c = 0.866 \ c

So the distance appeared to Charlie due to the length contraction,

$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$

$L_b=10\times \sqrt{1-0.75} \ Ly$

    $=5 \ Ly$

The time required to finish the race by Charlie is

$t_b = \frac{L_c}{V_c}$

  $=\frac{5 \ c}{0.866 \ c}$

  = 5.77 year

5 0
3 years ago
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