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ASHA 777 [7]
3 years ago
8

Use the circuit to answer the following questions.

Physics
1 answer:
Zanzabum3 years ago
7 0

Answer:

1)ammeter

2)ised to check measure of current flow through a circuit

3)o.90 ambere

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If 40 kJ of heat energy are added to 4 kg of water at 30°C, what will be the final temperature of the water? The specific heat o
n200080 [17]

Heat gained or added in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)
40000 = 4(4186)(T2 - 30.0)
<span>T2 = 32.4 degrees Celsius</span></span>

5 0
3 years ago
A 1500 kg car traveling at 8 m/s slows to a stop when the driver hits the brakes. If the braking force between the tires and the
wariber [46]

Answer:

First you fond the total force the car initialy has which is F=ma so it is 1500 times 8 which leads you to get 12000N then you divide the force of the car by the breaks and the road (4200N) which gives you 2.85 seconds for the car to come to a stop.

5 0
4 years ago
The latitude of the city of Arlington is about 32.7357°. Calculate the following: (a) The angular speed of the Earth. (b) The li
julia-pushkina [17]

Answer:

Explanation:

a ) The earth rotates by 2π radian in 24 x 60 x 60 s

so angular speed ( w )  = 2π / (24 x 60 x 60)  = 7.268 x  10⁻⁵ rad / s

b ) Linear speed of city of Arlington ( v )  = w r = w R Cosλ where R is radius of the earth and λ is latitude .

v = 7.268 x 10⁻⁵ x  6.371 x 10⁶ cos 32.7357

389.5 m /s

acceleration = w² r = w² R Cos 32.7357

= (7.268 x 10⁻⁵ )² x 6.371 x 10⁶ x cos 32.7357

=283.08 x 10⁻⁴ m/s²

c) velocity ratio =

w r /w R =

R cos 32.73/ R  

= Cos 32.73

= 0.84 .

6 0
4 years ago
Which part of a circuit can be turn on and off to make an electromagnet work or stop
Charra [1.4K]

Answer:The electric current

Explanation:

8 0
3 years ago
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti
Ugo [173]

Use Newton's second law to determine the acceleration being applied to the sled. There are three forces at work on the sled (its weight, the force normal to the ground, and friction) but two of them cancel, leaving friction as the only effective force. This vector is pointed in the opposite direction of the sled's movement, so if we take the direction of its movement to be the positive axis, we would find the acceleration due to the friction to be

\vec F_G+\vec F_N+\vec F_F=m\vec a\iff-12\,\mathrm N=(20\,\mathrm{kg})a\implies a=-0.6\,\dfrac{\rm m}{\mathrm s^2}

Now we use the formula

{v_f}^2-{v_i}^2=2a(x_f-x_i)

to find the distance it travels. The sled comes to a rest, so v_f=0, and let's take the starting position x_i=0 to be the origin. Then the distance traveled x_f-x_i=x_f is

-\left(4.5\,\dfrac{\rm m}{\rm s}\right)^2=2\left(-0.6\,\dfrac{\rm m}{\mathrm s^2}\right)x_f\implies x_f\approx17\,\mathrm m

6 0
3 years ago
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