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3 years ago

What 2 aspects of a force do scientists measure???

Physics

2 answers:

sdas [7]3 years ago

7 0

Answer: magnitude and direction

Explanation:They are the two aspects of force that scientists measure

sveticcg [70]3 years ago

6 0

Answer:

Explanation:

Force is a VECTOR QUANTITY which means it is DEFINED by its MAGNITUDE AND DIRECTION

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What is true about colonial families?

MrRissso [65]

Answer:

Explanation:

I just took the test and C was wrong, do not listen to that answer. It is "All of the above"

7 0

3 years ago

A model car is 29 cm in length. How many inches long is it?

Delvig [45]

1 cm = 0.39 in.

29 cm = 0.39 * 29

29 cm = 11.41 in.

7 0

4 years ago

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A bullet is shot from a rifle with a speed of 3,015 feet per second. Assuming the billet moves at a constant velocity what is th

VMariaS [17]

Answer:

Explanation:

Given

final speed v = 3015ft/s

initial speed = 0m/s

Distance S = 4146m/s

Required

Time

Using the equation of motion to get the time;

S = (v+u)/2 * t

since 1m = 3.28084ft'

4146m = 3.28084 * 4146 = 13,602.36264feet

13,602.36264 = 3015+0/2 * t

13,602.36264 = 1,507.5t

t = 13,602.36264/ 1,507.5

t = 9.023 secs

Hence it will take 9.023 seconds the rescuer die the biller to strike a target

3 0

3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

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3 years ago

True or False?

WINSTONCH [101]

Answer:

true is the answer of the question

5 0

3 years ago

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