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ludmilkaskok [199]

3 years ago

10

Georgie was pulling her brother (of mass 16 kg) in a 9.4 kg sled with a constant force of 39 N for one block (128 m). How much w

ork did Georgie do? Answer in units of J.

1 answer:

Genrish500 [490]3 years ago

7 0

Answer:

4992 J

Explanation:

Force, F = 39 N

Distance, s = 128 m

Work done = force x distance

W = 39 x 128

W = 4992 J

Thus, the work done is 4992 J.

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A projectile is fired with an initial speed of 60.3 m/s at an angle of 34.2 above the horizontal on a long flat firing range.

amid [387]

Answer:

A.) H = 58.6 m

B.) T = 6.92 s

C.) 345.12 m

D.) V = 22.13 m/s

E.) Ø = 32.1 degree

Explanation:

Given that the

initial speed U = 60.3 m/s

Angle Ø = 34.2 degree

A.) At maximum height, final velocity V is equal to zero.

Using the third equation of motion under gravity.

V^2 = U sin Ø^2 - 2gH

Substitute for U and g. Where g = 9.8 m/s^2

0 = (60.3 sin 34.2)^2 - 2 × 9.8 × H

1148.78 = 19.6 H

H = 1148.78/19.6

H = 58.6 m

B.) To Determine the total time in the air, let us use the formula

V = UsinØ - gt

At maximum height, V = 0

t = UsinØ/g

Total time T = 2t

Therefore, T = 2UsinØ/g

T = (2 × 60.3 × sin 34.2)/9.8

T = 67.79/9.8

T = 6.92 s

C.) To determine the total horizontal distance covered which is the range, we will use second equation of motion.

S = UcosØT - 1/2gt^2

Where S = range R

g = 0, since the range is not a vertical distance

T = total time

Substitute all the parameters into the formula

R = 60.3 cos 34.2 × 6.92

R = 345.12 m

D.) After 1.2 s firing,

V = UsinØ - gt

Where t = 1.2 s

Substitute into the formula

V = 60.3 × sin34.2 - 9.8 × 1.2

V = 33.89 - 11.76

V = 22.13 m/s

Therefore the speed of the projectile 1.20 s after firing is 22.13 m/s

E.) The direction will be determined by using the formula

t = VsinØ/ g

Cross multiply

VsinØ = gt

Make SinØ the subject of formula

SinØ = gt/V

SinØ = (9.8×1.2)/22.13

Sin Ø = 11.76/22.13

Sin Ø = 0.53

Ø = sin^-1( 0.53 )

Ø = 32.1 degree

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3 years ago

A 332 kg mako shark is moving in the positive direction at a constant velocity of 2.30 m/s along the bottom of a sea when it enc

Digiron [165]

To solve this problem we will apply the concepts related to the conservation of momentum. By definition we know that the initial moment must be equivalent to the final moment of the two objects therefore

$p_1 = p_2$

$m_1u_1+m_2u_2 = m_1v_1+ m_2v_2$

Here,

$m_{1,2} =$ Mass of each object

$u_{1,2} =$ Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

Since the initial velocity relative to the metal tank is at rest, that velocity will be zero. And considering that in the end, the speed of the two bodies is the same, the equation would become

$m_1u_1 = (m_1+m_2)v_f$

Rearranging to find the velocity,

$v_f = \frac{m_1u_1}{ (m_1+m_2)}$

Replacing we have that,

$v_f = \frac{(332)(2.3)}{ (332+19.5)}$

$v_f = 2.17 m/s$

Therefore the velocity of the shark immediately after it swallows the tank is $2.17m/s$

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3 years ago

A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20.0° with the horizontal. Using t

Elis [28]

The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2

Solve for <em>v</em> :

<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

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The airplanes acceleration is 2.1875

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As an object fall in a gravitational field, its speed increases. This is an example of potential energy transforming into what?

IrinaK [193]

Answer:

potential energy is transformed into kinetic energy

5 0

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