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3 years ago

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A soccer player kicks a soccer ball with a force of 1.8 N. If the mass of the ball is .43 kg. How fast will the ball accelerate?

2 answers:

Vesnalui [34]3 years ago

4 0

Answer:

The acceleration of the ball is 4.18 [m/s^2]

Explanation:

By Newton's second law we can find the acceleration of the ball

$F = m*a\\where:\\F = force applied [N] or [kg*m/s^2]\\m = mass of the ball [kg]\\a = acceleration [m/s^s]$

Now we have:

$a = F/m\\a = \frac{1.8 [kg*m/s^s]}{0.43[kg]} \\a = 4.18 [kg]$

saw5 [17]3 years ago

3 0

Answer: The acceleration of the ball is a = 4.186m/s^2

Explanation: By the second Newton's Law we have that:

F = m*a

where F is force, m is mass and a is accleration.

In this case, we have F = 1.8N and m = 0.43kg, so we have to solve the equation:

1.8N = 0.43kg*a

a = 1.8N/0.43kg = 4.186m/s^2

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A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m . What is the effective spring

ANEK [815]

Answer:

29856.521 N/m

Explanation:

$m=Mass\ of\ diver=70\ kg\\x=Length\ compressed\ by\ spring=2.3\times 10^{-2}\ m\\a=Acceleration\ due\ to\ gravity=9.81\ m/s^2\\F=Force\ exerted\ by\ diver=m\times a\\\Rightarrow F=70\times 9.81\\\Rightarrow F=686.7\ N\\k=spring\ constant\\F=k\times x\\\Rightarrow k=\frac {F}{x}\\\Rightarrow k=\frac {686.7}{2.3\times 10^{-2}}\\\therefore k=29856.521\ N/m$

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What happens to centripetal acceleration as the radius of curvature decreases and the speed is constant, and why

lutik1710 [3]

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Hopr it helps :)

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3 years ago

A car travels west for 240 km in 4 h. what is the car's velocity?

nalin [4]

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3 years ago

Read 2 more answers

A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.

ValentinkaMS [17]

A) $2.03 m/s^2$

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

$mg sin \theta - \mu_k R = ma$ (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

$\theta=22.5^{\circ}$

$\mu_k = 0.19$ is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

$R-mg cos \theta =0$ (2)

From (2) we find

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_k mg cos \theta = ma$

Solving for a,

$a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2$

B) 5.94 m/s

We can solve this part by using the suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

$a=2.03 m/s^2$

s = 8.70 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s$

6 0

3 years ago

Please help ASAP! Thank you :)

puteri [66]

Answer:

magnitude of gravitational force between the Earth and the Sun at B is greater than that at A

Explanation:

Formula of gravitational force:

F = GMm/r^2

(r is the distance between 2 objects)

We see that r(B) < r(A) since at B, the Earth is closer to the Sun than at A

According to the Formula, the smaller r is, the greater F is

So, F(B) > F(A)

8 0

3 years ago