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3 years ago

The unit for measuring electric power is the A. ampere. B. volt. C. ohm. D. watt.

1 answer:

6 0

The correct answer is D: Watt. This unit was named after James Watt, and is used to express the equivalent of one joule per second in energy. In experiments and on the packaging for electrical products such as light-bulbs, the measurement will usually be written in its abbreviated format: W.

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What is the formula that represents the simplest ratio of the atoms in the components

Dafna1 [17]

Answer:

Empirical formula

Hope it'll help!

stay safe:)

5 0

3 years ago

The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of 216 km/h. (a) If the train goes

faust18 [17]

Answer:

a) r = 6122 m and b) v = 32.5 m / s

Explanation:

a) The train in the curve is subject to centripetal acceleration

a = v2 / r

Where v is The speed and r the radius of the curve

They indicate that the maximum acceleration of the person is 0.060g,

a = 0.060 g

a = 0.060 9.8

a = 0.588 m /s²

Let's calculate the radius

v = 216 km / h (1000m / 1km) (1 h / 3600 s =

v = 60 m / s

r = v² / a

r = 60² /0.588

r = 6122 m

b) Let's calculate the speed, for a radius curve 1.80 km = 1800 m

v = √a r

v = √( 0.588 1800)

v = 32.5 m / s

6 0

3 years ago

All of the following are functions of the skeletal system EXCEPT:

sasho [114]

I'm pretty sure the answer would be D

8 0

3 years ago

Read 2 more answers

Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k

Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is $\bf{6.59~m/s}$ .

Part(B): The direction is $\bf{42.7^{0}}$ south to east.

Explanation:

Given:

Mass of Jack, $m_{1} = 59.0~Kg$

Mass of Jill, $m_{2} = 47..0~Kg$

Initial velocity of Jack, $v_{1i} = 8.00~m/s$

Initial velocity of Jill, $v_{2i} = 0$

Final velocity of Jack, $v_{1f} 5.00~m/s$

The final angle made by Jack after collision, $\alpha = 34.0^{0}$

Consider that the final velocity of Jill be $v_{2f}$ and it makes an angle of $\beta$ with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

$~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}$

where, $v_{2f}^{x}$ is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

$v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s$

Conservation of momentum of the system along north direction is given by

$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

where, $v_{2f}^{y}$ is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

$v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s$

Part(B):

The direction is given by

$\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}$

4 0

3 years ago

A woman on a bridge 95.6 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an a

NARA [144]

Answer:

The speed of the raft is 1.05 m/s

Explanation:

The equation for the position of the stone is as follows:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the stone at time t

y0 = initial height

v0 = initial speed

t = time

g = acceleration due to gravity

The equation for the position of the raft is as follows:

x = x0 + v · t

Where:

x = position of the raft at time t

x0 = initial position

v = velocity

t = time

To find the speed of the raft, we have to know how much time the raft traveled until the stone reached the river. For that, we can calculate the time of free fall of the stone:

y = y0 + v0 · t + 1/2 · g · t² (v0=0 because the stone is dropped from rest)

If we place the origin of the frame of reference at the river below the bridge:

0 m = 95.6 m - 9.8 m/s² · t²

-95.6 m / -9,8 m/s² = t²

t = 3.12 s

We know that the raft traveled (4.84 m - 1.56 m) 3.28 m in that time, then the velocity of the raft will be:

x/t = v

3.28 m / 3.12 s = v

v = 1.05 m/s

5 0

2 years ago

Read 2 more answers