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3 years ago

A 200 N force is applied to a 50.0 kg object moving at 3.0 m/s. How fast will it be going at 4.0 s?

Physics

1 answer:

blsea [12.9K]3 years ago

8 0

Answer:

19.0 m/s

Explanation:

First of all, we can find the acceleration of the object by using Newton's second law:

$F=ma$

where

F = 200 N is the net force applied to the object

m = 50.0 kg is the mass of the object

a is the acceleration

Solving for a,

$a=\frac{F}{m}=\frac{200 N}{50.0 kg}=4.0 m/s^2$

Now we can find the final speed of the object:

$v=u+at$

where

u = 3.0 m/s is the initial speed

a = 4.0 m/s^2 is the acceleration

t = 4.0 s is the time

Substituting,

$v=3.0 m/s + (4.0 m/s^2)(4.0 s)=19.0 m/s$

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3 years ago

Read 2 more answers

6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t

Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of 12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

$y = u sin \theta - \dfrac{1}{2}gt^2$ ................(2)

from equation(1) and (2)

$y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}$ ..........{3}

$12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}$

$\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29$

$\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2$

$u = \sqrt{10959.34}$

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

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3 years ago

What are the names of neptunes moons

alexandr402 [8]

Answer:

1. naiad

2.Thalassa

3 Despina

4 Galatea

5 Larissa

6 Hippocamp

7 Proteus ˈ

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Explanation:

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5 0

3 years ago

Compute the dot product of the vectors u and v, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and

OLga [1]

Answer:

$\theta = 106.3 degree$

Explanation:

As we know that

$\vec w = -\hat i + 7\hat j$

$\vec v = 7\hat i - \hat j$

also we know that

$\vec v. \vec w = -14$

it is given as

$\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)$

$\vec v. \vec w = - 7 - 7 = -14$

also we can find the magnitude of two vectors as

$|v| = \sqrt{(-1)^2 + (7)^2}$

$|v| = \sqrt{50}$

similarly we have

$|w| = \sqrt{(7^2) + (-1)^2}$

$|w| = \sqrt{50}$

now we know the formula of dot product as

$\vec v. \vec w = |v||w| cos\theta$

$-14 = (\sqrt{50})^2cos\theta$

$\theta = cos^{-1}(\frac{-14}{50})$

$\theta = 106.3 degree$

3 0

3 years ago

A 3500 N is force is applied to a spring that has a spring of constant of k= 14000 N/m. How far from equilibrium will the spring

Taya2010 [7]

Answer:

the spring be displaced by 25.0 cm

Explanation:

The computation is shown below:

As we know that

F= -K × x

So,

$x = \frac{-F}{K}$

Now

$x = \frac{-3500}{14000} \\\\$

= -0.250m

= 25.0 cm

Hence, the spring be displaced by 25.0 cm

4 0

3 years ago