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levacccp [35]
3 years ago
14

What is the mass of a basketball that has Connecticut energy of 102 J and it's travelling at 5 m/s

Physics
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

8.16Kg

Explanation:

K.E= 1/2 mv²

102=1/2 m 5²

m=8.16 Kg

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Soccer was made in 19th December 1863, football was November 6, 1869 (ha), baseball was 1839, track and field is 776 BC, 
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Ohm's law states which relationship between electrical quantities? a. Volts = current x resistance b. Volts = current divided by
miv72 [106K]

Answer:

a. Volts = current x resistance

Explanation:

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

V = IR

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

Hence, Ohm's law gives the relationship between voltage, current and resistance of an electric circuit.

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1-5 answer please .....
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5). Free body diagram
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3 years ago
When a charge of 8 C flows past any point along a circuit in 2 seconds, the current is ________ A?
RoseWind [281]
The answer would be 4 A.
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3 years ago
Suppose that the hatch on the side of a Mars lander is built and tested on Earth so that the internal pressure just balances the
Kaylis [27]

Answer:

The value is  F_{net} =  4444 lb

The force will be  outward

Explanation:

From the question we are told that

   The diameter of the disk is  d = 50.0 \ cm  = \frac{50}{100} = 0.5 \ m

   The external pressure on Mars  is P = 650 \ N/m^2

From the question we are told that  

   Internal pressure  =  External pressure

Generally the external Force on earth is

      F_E = P_{atm} * A

Here P_{atm} is the atmospheric pressure with value  P_{atm} = 1.013*10^{5}\ Pa

So

      F_E = 1.013 *10^{5} * \pi * \frac{d^2}{4}

=>   F_E = 1.013 *10^{5} *3.142 * \frac{0.50 ^2}{4}

=>   F_E = 19893 \  N

Generally the external Force on Mars is  

       F= P * A

      F = 650 * \pi * \frac{d^2}{4}

=>   F = 650 *3.142 * \frac{0.5^2}{4}

=>   F = 127.6 \  N    

Net force is mathematically represented as

      F_{net} = F_E -F

=>    F_{net} =  19893  -127.6

=>    F_{net} =  19765.6 \ Nconverting to  pounds

    F_{net} = \frac{19765.6}{4.448}

=> F_{net} =  4444 lb

Given that that the value is positive then the force will be  outward

7 0
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