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Damm [24]
3 years ago
14

If C is the vector sum of A and B, C = A + B, What must be true if C = A + B?. What must be true if C = 0?.

Physics
1 answer:
dybincka [34]3 years ago
7 0
A and B are equal vectors
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The answer is A. an object that allows some light to pass through
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2 years ago
1. A base-ball of mass 0.3kg approaches the bat at a speed of 30 miles/hour and when the ball hits the bat for 0.5 s, it started
andre [41]

Answer:

I = 27kg.mi/h

Explanation:

In order to calculate the impulse of the ball, you use the following formula:

I=m\Delta v  =m(v-v_o)      (1)

m: mass of the ball = 0.3kg

v: speed of the ball after the bat hit it = 60mi/h

vo: speed of the ball before the bat hit it = 30mi/h

You replace the values of all parameters in the equation (1):

I=(0.3kg)(60mi/h-(-30mi/h))=27kg\frac{mi}{h}

where the minus sign of the initial velocity means that the motion of the ball is opposite to the final direction of such a motion.

The imulpse of the ball is 27 kg.miles/hour

5 0
3 years ago
A solid must be given​
alexdok [17]
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8 0
3 years ago
A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed
gregori [183]

Answer:

the rate of the change of the length of the shadow is - 0.8625 m/s.

The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.

Explanation:

Given the data in the question;

Let x represent the man's distance from building,

initially x = 1m2

dx/d t= -2.3 m/s

Also Let y represent shadow height

so we determine dy/dt when x is 4m from the building

form the image description of the problem, we see two-like triangles with the same base and height ratios

so

 2 / (12-x) = y / 12

24 = y(12 - x )

y = 24 / (12-x)

dy/dt = 24/(12-x)² × dx/dt

Now at x = 4,

we substitute

dy/dt will be;

⇒ 24/(12 - 4)² × -2.3

= 24/64 - 2.3

= 0.375 × -2.3

dy/dt = - 0.8625 m/s

Therefore, the rate of the change of the length of the shadow is - 0.8625 m/s.

The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.

5 0
3 years ago
At the outer edge of a rotating space habitat, 130 m from the center, the rotational acceleration is g. What is the rotational a
enyata [817]

Answer:

Explanation:

Given:

R1 = 130 m

R2 = 65 m

w^2R = g

Assume, g = 9.81 m/s^2

w^2 = 9.81/130

w = 0.275 rad/s

At R2 = 65 m

g = w^2R

= (0.275^2) × 65

= 4.905 m/s^2

In conclusion,

g × R = k

g1/R1 = g2/R2

g2 = (g1 × 65)/130

= g1 ×1/2

= g1/2

6 0
3 years ago
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