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3 years ago

How can we measure electromotive force of a battery illustrated answer with the help of circuit diagram

Physics

1 answer:

MaRussiya [10]3 years ago

7 0

Answer:

the emf of the cell can be determined by measuring the voltage across the cell using a voltmeter and the current in the circuit using an ammeter for various resistances.

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How do l calculate e<br> nergy

aivan3 [116]

The formula for energy of motion is KE = .5 x m x v^2

Ke= Kinetic Energy in Joules

m = Mass in Kilograms

v = Velocity in Meters per Second

4 0

2 years ago

Scientists want to place a telescope on the moon to improve their view of distant planets. The telescope weighs 200 pounds on Ea

LuckyWell [14K]

Answer:

Explanation:

the weight of the telescope decreases because the moon attract the body with less force as compared to earth due to less gravity as compared to earth

6 0

3 years ago

A particle released during the fission of uranium-235 is a(n) ________

Kaylis [27]

<span>A particle released during the fission of uranium-235 is a "Neutron"</span>

6 0

2 years ago

The density of a hippo is approximately 1030kg/m^3,so it sinks to the bottom of the freshwater lakes and rivers. A 1500kg hippo

hram777 [196]

Answer:

14,700 N

Explanation:

The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:

$W-N=0$ (1)

the weight of the hyppo is

$W=mg=(1500 kg)(9.8 m/s^2)=14,700 N$

where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find

$N=W=14,700 N$

8 0

3 years ago

A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the

Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

v = u + at

0 = 19.09 - 9.81 t

t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

S = ut + 0.5 at²

H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

Maximum height of the ball = 18.57 m

6 0

2 years ago