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IceJOKER [234]
2 years ago
11

a ball is tossed straight up in the air to a maximum height of 25 meters before returning to the point that it was thrown, where

it is caught by the one who threw it. When the ball is initially thrown, the thrower imparts an angular velocity of 8 rad/s on the ball, causing it to spin. How many revolutions does the ball make before the person catches the ball?
Physics
1 answer:
boyakko [2]2 years ago
5 0

Answer:

height=25

velocity=8

force of gravity=10

Answer =2000

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An engine moves a motorboat through water at a constant velocity of 22 meters/second. If the force exerted by the motor on the b
trapecia [35]

Answer: Option B: 1.3×10⁵ W

Explanation:

Power = \frac{Work \hspace{1mm} done}{Time}

P=\frac{W}{t}

Work Done, W= F.s

Where s is displacement in the direction of force and F is force.

\Rightarrow P = \frac{F.s}{t} =F \times \frac{s}{t}=F.v

where, v is the velocity.

It is given that, F = 5.75 × 10³N

v = 22 m/s

P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W

Thus, the correct option is B

6 0
3 years ago
Can a body have constant speed and still be acclerating ? Give an Example
Allushta [10]






Hi Pupil Here's Your answer :::





➡➡➡➡➡➡➡➡➡➡➡➡➡





An object moving with constant speed can be accelerated if direction of motion changes. For example, an object moving with a constant speed in a circular path has an acceleration because its direction of motion changes continuously.





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3 0
4 years ago
An ideal battery would produce an extraordinarily large current if "shorted" by connecting the positive and negative terminals w
liubo4ka [24]

Answer:

Resistance, R=0.529\ \Omega

Explanation:

Given that,

Voltage of the battery, V = 9 volts

Current produced in the circuit, I = 17 A

We need to find the resistance when shorted by a wire of negligible resistance. It is a case of Ohm's law. The voltage is given by :

V=IR

R=\dfrac{V}{I}

R=\dfrac{9\ V}{17\ A}

R=0.529\ \Omega

So, the resistance in the circuit is 0.529 ohms. Hence, this is the required solution.

6 0
3 years ago
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
3 years ago
Read 2 more answers
Three systems of two charged particles are shown below. All the particles have the same mass,
Len [333]

Answer:D

Explanation:

5 0
3 years ago
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