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3 years ago

5

the strength of the pulling force on the moon is 1/6 of that on earth whilst on Jupiter it is 2.5 times stronger what is the wei

ght on the moon

1 answer:

KiRa [710]3 years ago

6 0

Weight is the force exerted on an object by gravity. So, weight of any object on the moon is 1/6 that on Earth.

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A soccer ball is kicked with an initial velocity 13 m/s at an angle of 35 degrees from the horizontal. What is the maximum heigh

Nana76 [90]

Answer:

The maximum height reached by the ball is 2.84 m

Explanation:

Given;

initial velocity of the soccer, u = 13 m/s

angle of projection, θ = 35°

The maximum height reached by the ball = ?

Apply the following kinematic equation, to determine the maximum height reached by the ball.

Maximum height (H) is given as;

$H = \frac{U^2sin^2\theta}{2g} \\\\H = \frac{(13)^2(sin \ 35^0)^2}{2*9.8}\\\\H = 2.84 \ m$

Therefore, the maximum height reached by the ball is 2.84 m

3 0

3 years ago

A technician is troubleshooting a problem. The technician tests the theory and determines the theory is confirmed. Which of the

vladimir1956 [14]

Answer:

b) Document lessons learned.

Explanation:

First he should do documentation

then C

then D

then A

3 0

3 years ago

A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.4 m/

Bad White [126]

Answer:

1.06 secs

Explanation:

Initial speed of sled, u = 8.4 m/s

Final speed of sled, v = 5.8 m/s

Coefficient of kinetic friction, μ = 0.25

Using the impulse momentum theory, we know that the impulse applied to the sled is equal to change in momentum of the sled:

FΔt = mv - mu

where m = mass of the object

Δt = time interval

F = force applied

The force applied on the sled is the frictional force, which is given as:

F = -μmg

where g = acceleration due to gravity

Therefore:

-μmgΔt = mv - mu

-μmgΔt = m(v - u)

-μgΔt = v - u

Making Δt subject of formula:

Δt = (v - u) / -μg

Δt = (5.8 - 8.4) / (-0.25 * 9.8)

Δt = -2.6/ -2.45

Δt = 1.06 secs

It took the sled 1.06 secs to travel from A to B.

7 0

3 years ago

Kara Less was applying her makeup when she drove into South's busy parking lot last Friday morning. Unaware that Lisa Ford was s

exis [7]

Answer

given,

Mass of Kara's car = 1300 Kg

moving with speed = 11 m/s

time taken to stop = 0.14 s

final velocity = 0 m/s

distance between Lisa ford and Kara's car = 30 m

a) change in momentum of Kara's car

Δ P = m Δ v

$\Delta P = m (v_f-v_i)$

$\Delta P = 1300 (0 - 11)$

Δ P = - 1.43 x 10⁴ kg.m/s

b) impulse is equal to change in momentum of the car

I = - 1.43 x 10⁴ kg.m/s

c) magnitude of force experienced by Kara

I = F x t

I is impulse acting on the car

t is time

- 1.43 x 10⁴= F x 0.14

F = -1.021 x 10⁵ N

negative sign represents the direction of force

8 0

3 years ago

Read 2 more answers

What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of 85 km/h so that the occ

borishaifa [10]

Answer:

k = 5178.8 N/m

Explanation:

As we know that spring mass system will oscillate at angular frequency given as

$\omega = \sqrt{\frac{k}{m}}$

now we have

$\omega = \sqrt{\frac{k}{1200}}$

now the maximum acceleration of the spring block system is at its maximum compression state which is given as

$a = \omega^2 A$

here A= maximum compression of the spring

so here in order to find maximum compression of the spring we will use energy conservation as we know that initial total kinetic energy of the car will convert into spring potential energy

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2$

here we know that

v = 85 km/h

$v = 85 \times\frac{1000}{3600} = 23.61 m/s$

now we have

$(1200)(23.61^2) = kA^2$

$A^2 = \frac{6.68 \times 10^5}{k}$

now from above equation of acceleration we have

$5.0 g = (\frac{k}{m})\sqrt{\frac{6.68 \times 10^5}{k}}$

$5.0(9.81) = \sqrt{k}(0.68)$

$k = 5178.8 N/m$

6 0

3 years ago