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3 years ago

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A force of 200N is being applied over an area measuring 0.75m^2

1 answer:

Sonbull [250]3 years ago

3 0

Answer:

357KG

Explanation:

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How are newtons third law of motion applies to a system of objects

11111nata11111 [884]

Answer:

Newton's third law of motion states that whenever a first object exerts a force on a second object, the first object experiences a force equal in magnitude but opposite in direction to the force that it exerts. ... Newton's third law is useful for figuring out which forces are external to a system.

Explanation:

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3 years ago

Using calcium’s atomic structure, as shown in the image, what is this element’s atomic number?

ElenaW [278]

Answer:

Atomic number of calcium is 20.

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3 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

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3 years ago

E. weight = mass x gravitational field strength

allsm [11]

Answer:

e. weight = mass x gravitational field strength

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2 years ago

A player at first base catches a throw traveling 38 m/s. The baseball, which has a mass of 0.145 kg, comes to a complete stop in

mixas84 [53]

Answer:

F = 39.36 N

Explanation:

given,

initial speed, u = 38 m/s

final speed, v = 0 m/s

mass of ball = 0.145 Kg

time, t = 0.14 s

Force = ?

using impulse formula

J = change in momentum

J = F x t

m(v - u) = F x t

0.145 x (0 - (-38)) = F x 0.14

F x 0.14 = 5.51

F = 39.36 N

force exerted by the ball is equal to 39.36 N.

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3 years ago