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Leya [2.2K]
3 years ago
15

Please help on this one? PLEASE.

Physics
1 answer:
Fantom [35]3 years ago
6 0

Answer: The answer is D

Explanation:

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An object is thrown with velocity 7.1 m s-1 vertically upwards on the Moon. The
dem82 [27]

Answer:D

Explanation:

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2 years ago
A battery derives electric energy from _____energy?
sammy [17]
A). nuclear
No.  There were batteries long long before we learned
how to use nuclear energy.  Also, there is no danger of
exposure to radioactivity when you're working with a battery.

b). mechanical
No.  A battery has no moving parts.

c). gravitational
No.  No matter how high you take a battery in an airplane, or
how far you lower it into a mine-shaft, its characteristics don't
change.  In fact, batteries even work on things that are in orbit.

d). chemical
Bingo.
4 0
3 years ago
Read 2 more answers
1) Write Newton's three laws of motion (20 points). 2) Define acceleration (20 points). 3) Write the Month and Day of the Assign
vaieri [72.5K]

Answer:

Hey

I have no idea when YOUR assignment is due.

Newtons 1rst law:

An object that has constant motion will remain at that speed unless acted on by an external force.

Newtons 2nd law:

F=ma (force=mass*acceleration)

Newtons 3rd law:

when a force is applied to an object, there will be an opposite but equal reaction.

Acceleration:

How much your speed increases/decreases per unit of time.

I wrote all that^

4 0
2 years ago
A wire that is 0.86 meters long is moved perpendicularly through a constant magnetic field of strength 0.035 newtons/amp·meter a
GaryK [48]
The emf induced = B*l*v where B is the flux density, l the length of the conductor and v the velocity of the conductor. In the given case B = 0.035 N/amp.meter, l = 0.86 and v = 6 m/sec
emf = 0.035*0.86*6 = 0.1806 v ≈ 0.18 v
choice: D
6 0
3 years ago
A 10 N board of uniform density is 5 meters long. It is supported on the left by a string bearing a 3 N upward force. In order t
NikAS [45]

Answer:

C.\frac{25}{7}m

Explanation:

We are given that

Weight of board=w=10 N

Length of board=L=5 m

Tension in the string=T=3 N

Applied upward force=F=7 N

We have to find the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.

Let r be the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.

The board is uniform therefore, the center of board is the mid- point of board.

Therefore, the lever arm of weight=r_1=\frac{L}{2}=\frac{5}{2}m

Now, the torque exerted by the weight of the board

\tau_1=Force\times perpendicular\;distance=10\times \frac{5}{2}=25 N

The torque exerted  by applied force=\tau_2=7\times r=7r

In static equilibrium

The sum of rotational forces=0

\tau_1+\tau_2=0

The two rotational force act in opposite direction therefore,

\tau_2=-7r

Substitute the values

25-7r=0

7r=25

r=\frac{25}{7}m

Hence, option C is true.

7 0
3 years ago
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