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maxonik [38]
4 years ago
5

Melanie watched the path a baseball followed after a pitcher threw it. She noticed that the ball traveled horizontally away from

the pitcher as well as downward toward the ground. What force caused the ball to accelerate in the downward direction when it was thrown?
Physics
1 answer:
Alex17521 [72]4 years ago
7 0

Answer:gravity

Explanation:

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Your portable phone's 12V battery is dead. It's a holiday and all the stores are closed. You do have two 1.5V batteries from you
Anastaziya [24]

Answer:

No.

Explanation:

A transformer requires AC to work. A battery delivers DC only.

3 0
4 years ago
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A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx3 , where c = 8.0 N/m3 . The pa
Pie

Answer:4.58 m/s

Explanation:

Given

mass of Particle m=4 kg

F=-cx^3

a=\frac{F}{m}

a=-\frac{cx^3}{m}

a=-\frac{8x^3}{4}

a=-2x^3

v\frac{\mathrm{d} v}{\mathrm{d} x}=-2x^3

vdv=-2x^3dx

integrating

\int_{6}^{v_b}vdv=\int_{1}^{-2}-2x^3dx

\frac{v_b^2-6^2}{2}=-\frac{1}{2}\left [ \left ( -2\right )^4-\left ( 1\right )^4\right ]

\frac{v_b^2-36}{2}=-0.5\times 15

v_b^2=36-15

v_b=\sqrt{21}

v_b=4.58 m/s

6 0
3 years ago
A bug is moving along a meter stick in the negative direction at a constant speed of 0.85cm/s. After 42 seconds have passed the
Agata [3.3K]

Given that,

bug speed, v= 0.85 m/s

time, t =42 s

Final position of bug on meter stick was 27 cm

Starting position of bug on meter stick = ?

Since we know that,

s = vt

s= 0.85*42 = 35.7 cm

this is the distance covered by bug in the given time and velocity.

since the bug is moving in negative direction, starting point will be:

27.0 cm+ 35.7 cm = 62.7 cm

The bugs starting position on meter stick was 62.7 cm.

3 0
4 years ago
When charges qa, qb, and qc are placed respectively at the corners a, b, and c of a right triangle, the potential at the midpoin
earnstyle [38]

Answer:

8v

Explanation:

First we apply super position principle

Vt= v1 + v2+ v3

Remove qa

But vt= 20v

So V = v2+v3

V1= 20-15

= 5v

Remove qb

V= v1+v3

V=8v

So the potential when qa and qc are remove is the potential due to qb

Which is 8v

7 0
3 years ago
A voltmeter reads 200 V between parallel plates that are 2 cm apart. what is the strength of the electric field between the two
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I think that is c) l’m not sure
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3 years ago
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