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Harman [31]
3 years ago
9

A high jumper jumps over a bar that is 2 m above the mat. With what velocity does the jumper strike the mat in the landing area?

(Disregard air resistance. g = 9.81 m/s2.)
Physics
1 answer:
docker41 [41]3 years ago
8 0

Answer:

The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.

Explanation:

It is given that,

A high jumper jumps over a bar that is 2 m above the mat, h = 2 m

We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :

v=\sqrt{2gh}

g is acceleration due to gravity

v=\sqrt{2\times 9.81\ m/s^2\times 2\ m}

v = 6.26 m/s

So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.

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What is the voltage in a circuit if the current is 6.2 A and the resistance is 18 ohms?
Nitella [24]
We already know the formula:
Voltage = Current * Resistance
In the given question, there are numerous information's that are already given.
Current = 6.2 A
Resistance = 18 ohms
Then
Voltage = 6.2 * 18 Volts
             = 111.6 Volt
So, the voltage in the circuit will be 111.6 volts. I hope it helps you.
4 0
3 years ago
Does anyone knows the answer??​
Tamiku [17]

Answer:

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Explanation:

8 0
2 years ago
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.50 mm. A 25
torisob [31]

Answer:

The electric field is 16666.66 V/m.

The surface charge density is 1.474\times10^{-7}\ C/m^2

The capacitance is 4.484\times10^{-12}\ F

The charge on each plate is 112.1\times10^{-12}\ C

Explanation:

Given that,

Area =7.70 cm²

Distance = 1.50 mm

Potential difference = 25.0 V

Suppose  we find the electric field between the plates, the surface charge density, the capacitance and the charge on each plates.

We need to calculate the electric field

Using formula of electric field

E=\dfrac{V}{d}

Put the value into the formula

E=\dfrac{25.0}{1.50\times10^{-3}}

E=16666.66\ V/m

We need to calculate the charge density

Using formula of charge density

\sigma=E\times\epsilon_{0}

Put the value into the formula

\sigma=16666.66\times8.85\times10^{-12}

\sigma=1.474\times10^{-7}\ C/m^2

We need to calculate the capacitance

Using formula of capacitance

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times7.60\times10^{-4}}{1.50\times10^{-3}}

C=4.484\times10^{-12}\ F

We need to calculate the charge

Using formula of charge

q=CV

Put the value into the formula

q=4.484\times10^{-12}\times25.0

q=112.1\times10^{-12}\ C

Hence, The electric field is 16666.66 V/m.

The surface charge density is 1.474\times10^{-7}\ C/m^2

The capacitance is 4.484\times10^{-12}\ F

The charge on each plate is 112.1\times10^{-12}\ C

8 0
3 years ago
A wave on a string is described by Y(x, t) : 15.0 sin( nclS - 4rrt),, where x and y are in centimeters and / is in seconds. (a)
KIM [24]

Answer:

dsf

Explanation:

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4 0
3 years ago
A 0.10 kg baseball travelling at 40 m s−1 hits straight back to the pitcher at 55 m s−1. The contact time is 0.01 seconds. What
Ilia_Sergeevich [38]

Answer:

1.5 kgms⁻¹

Explanation:

Momentum can be defined as "<em>mass in motion</em>."  

The amount of momentum that an object has is dependent upon two factors

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  • speed of motiom 

when there is a change in the velocity , it creates a change in momentum also

when we consider that we can mathematically show this,In terms of an equation,

Change in momentum    (ΔΡ) = m(Δv)

where (Δv) - change in velocity

<em>(Δv) = final velocity - initial velocity</em>

Change in momentum (ΔΡ) = m(Δv)

                                             = 0.1×([55-40])

                                             = 1.5 kgms⁻¹

     

7 0
2 years ago
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