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lukranit [14]
2 years ago
7

What is the electric current measured in

Physics
1 answer:
cluponka [151]2 years ago
6 0

An ampere (AM-pir), or amp

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a crane does 9,500 J of work to lift a crate straight up using a force of 125 N. how high does the crane lift the crate?
Lina20 [59]

Answer:

<em>The crane lifts the crate up to 76 m high</em>

Explanation:

<u>Work Done by a Force</u>

The work done by a force of magnitude F that displaces an object by a distance y is given by

W=F.y

We know a crane does a work of 9,500 Joule to lift a crate and is using a force of 125 Nw.

The above equation can be solved to know the value of y in terms of the work and the force:

\displaystyle y=\frac{W}{F}

Plugging in the given values

\displaystyle y=\frac{9,500}{125}

y=76\ m

The crane lifts the crate up to 76 m high

7 0
3 years ago
*NEED ANSWER STAT*
tresset_1 [31]
Becomes a +1 ion for this
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3 years ago
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When dogs see police dogs do they think "oh no its the cops"?
ira [324]

No they say "Watch out it's the fuzz"

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3 years ago
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What is the relation between celsius and kelvin​
Scilla [17]

Answer:

The Celcius and kelvin scale are related unit for unit. One degree unit on the Celcius scale is equivalent to one degree unit on the kelvin scale. The only difference between these two scales is the zero point.

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3 years ago
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A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur
emmasim [6.3K]

• Net vertical force on the block:

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

(<em>n</em> = magnitude of normal force, <em>w</em> = weight)

<em>n</em> = <em>w</em> = <em>m g</em>

(<em>m</em> = mass, <em>g</em> = 9.8 m/s²)

<em>n</em> = (4 kg) (9.8 m/s²) = 39.2 N

• Net horizontal force:

∑ <em>F</em> = -<em>f</em> = <em>m a</em>

(<em>f</em> = mag. of friction, <em>a</em> = acceleration)

We have <em>f</em> = <em>µ</em> <em>n</em> = 0.5 (39.2 N) = 19.6 N, so

-19.6 N = (4 kg) <em>a</em>

<em>a</em> = -4.9 m/s²

With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆<em>x</em> in this time such that

0² - (5 m/s)² = 2 (-4.9 m/s²) ∆<em>x</em>

∆<em>x</em> = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m

8 0
3 years ago
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