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3 years ago

What is the acceleration of a 1.5 kg football thrown with a force of 13.00 N to the

Physics

1 answer:

nignag [31]3 years ago

5 0

Answer:

The acceleration of football is 6.00 m/s².

Explanation:

The acceleration can be calculated using the following equation:

$\Sigma F = ma$

Where:

F: is the force

m: is the mass

a: is the acceleration

$F_{e} - F_{w} = ma$

$13.00 N - 4.00 N = 1.5 kg*a$

$a = \frac{9.00 N}{1.5 kg} = 6.00 m/s^{2}$

Therefore, the acceleration of football is 6.00 m/s².

I hope it helps you!

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A ductile metal wire has resistance R. What will be the resistance of this wire in terms of R if it is stretched to three times

Ira Lisetskai [31]

Answer:

Explanation:

We know that the resistance is $R=\rho *\frac{L}{A}$ .

If we stretch the wire to a new length L2 = 3L, the cross-sectional area will also change. If the cross-sectional area doesn't change throughout the wire, we can say that:

Volume = L*A = 3L * A2 being A2 the new area after stretching the wire.

Since the volume remains the same we conclude that A2 = A/3

With this information, we calculate the new resistance:

$R2=\rho *\frac{L2}{A2}=\rho *\frac{3*L}{A/3}=\rho * 9 * \frac{L}{A}$

Since $R=\rho *\frac{L}{A}$ , and by simple inspection of the previous equation, we get:

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3 years ago

If 50 km thick crust having an average density of 3.0 g/cm3 has a surface elevation of 2.5 km above sea level, what would you pr

RUDIKE [14]

Answer:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

Explanation:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

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3 years ago

Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa

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