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2 years ago

5

You and your friend are having a discussion about weight. He claims that he weighs less on the on the 100th floor of a building

than he does on the ground floor. Is he correct? support your answer with evidence.

1 answer:

Andru [333]2 years ago

8 0

Answer:

SICENCE

Explanation: ITS RIGHT

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A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the v

Rasek [7]

Answer:

Explanation:

This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is

$[m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a$ Filling in:

$[1200(4.5)+2100(0)]=[(1200+2100)v]$ which simplifies to

5400 + 0 = 3300v

so v = 1.6 m/s to the east, choice B

6 0

2 years ago

Two rocks, A and B, are thrown horizontally from the top of a cliff. Rock A has an initial speed of 10 meters per second and roc

maria [59]

Answer:

i need help

Explanation:

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2 years ago

Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart

Romashka [77]

Answer:

E = 5291.00 N/C

Explanation:

Expression for capacitance is

$C = \frac{\epsilon A}{d}$

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

$C = \frac{\epsilon\times(19\times 10^{-2})^2}{1.5\times 10^{-3}}$

$C = 24.06 \epsilon$

$C = 24.06\times 8.854\times 10^{-12} F$

$C =2.1\times 10^{-10} F$

We know that capacitrnce and charge is related as

$V = \frac{Q}{C}$

$= \frac{2\tiimes 10^{-9}}{2.\times 10^{-10}}$

v = 9.523 V

Electric field is given as

$E = \frac{V}{d}$

= $\frac{9.52}{1.8*10^{-3}}$

E = 5291.00 V/m

E = 5291.00 N/C

5 0

2 years ago

A motorbike accelerates from 15m/s to 25m/s in 15 seconds.

RSB [31]

distance traveled by a uniformly accelerated bike is given as

$d = \frac{v_f + v_i}{2} (t)$

here we know that

$v_f = 25 m/s$

$v_i = 15 m/s$

$t = 15 s$

now we will have from above equation

$d = \frac{15 + 25}{2} (15)$

$d = 20 (15) = 300 m$

so it will cover the total distance of 300 m

5 0

3 years ago

Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of

FrozenT [24]

Gravity obeys the inverse square law. At 6400 km above the center of the Earth (Earth's surface) you weigh x. Twice that reduces your weight to 1/4th. Four times that height reduces your weight to 1/16th. 4 times 6400 km is 25,600 km. But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer: 19,200 km.

Incidentally, it doesn't exactly work the opposite way. At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.

6 0

3 years ago