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3 years ago

7

A battery-operated car moves forward as a result of which device? Electromagnet, generator, motor, or a transformer?

2 answers:

Romashka [77]3 years ago

7 0

A battery-operated car moves forward as a result of which device?

A) Electromagnet

B) Generator

<u>C) Motor </u>

D) Transformer

Natasha_Volkova [10]3 years ago

5 0

Answer:

a motor

Explanation:

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(PLS HELP 20 POINTS, IM TOO DUMB FOR THIS) What is the total momentum of the system after the collision?

Dmitry_Shevchenko [17]

Using the law of conservation of momentum
m1u1+m2u2=m1v1+m2v2
Where m1 is mass of first object
m2 is mass of second object
u1 and u2 are initial velocities of object 1 and 2 respectively
v1 and v2 are final velocities of object 1 and 2 respectively
Here, they are moving as a system after collision. Thus they will posses same final velocity
m1u1 +m2u2=v(m1+m2)
Substituting values
600*4+0=v(600+400)
2400=v*1000
v=2.4 m/s

Now momentum of system
p=Mv
p=(600+400)*2.4
p=1000*2.4
Therefore p=2400 kg m/s
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3 years ago

What is the symbol name?

murzikaleks [220]

Answer:

cell

Explanation:

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2 years ago

A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera

Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m= $2.1\times 10^3 kg$

Initial velocity,u=0

Distance,s=5.9 m

$\theta=19^{\circ}$

Average friction force,f= $4.0\times 10^3 N$

We have to find the speed of the car at the bottom of the driveway.

Net force, $F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3$

Where $g=9.8 m/s^2$

Acceleration, $a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}$

$v=\sqrt{2as}$

$v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}$

v=3.9 m/s

7 0

3 years ago

A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.

pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

$h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t = 6.55 \ s$

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

$X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m$

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

8 0

3 years ago

An oceanic ridge might be compared to what continental landform? A) delta B) great plains C) mountain range Eliminate D) mountai

Pavlova-9 [17]

A mountain range because an ocean ridge is an underwater mountain hope this helps you

6 0

3 years ago