A spaceship is accelerating at 1000 m/sec2 . How much force is required from the backthrusters to completely stop the spaceship?

Add me on my new account Conner_Bryant3 on brainly

Zarrin [17]

Yes fishnet cudhejeiei. Jen did she

5 0

3 years ago

The position of a ball is given by x(t) (2.3 m/s) .t + (5.3m/s3) t3. What is the position 348 m at time t 4 s? You are correct Y

Georgia [21]

Answer:

(a) 348.4 m

(b) 256.7 m/s

(c) 127.2 m/s^2

Explanation:

$x = 2.3 t + 5.3 t^{3}$

(a) at t = 4 s

x = 2.3 x 4 + 5.3 x 4 x 4 x 4

x = 348.4 m

(b) The derivative of displacement function gives the value of instantaneous velocity.

So, v = dx / dt = 2.3 + 5.3 x 3 x t^2

v = 2.3 + 15.9 t^2

Put t = 4 s

So, v = 2.3 + 15.9 x 4 x 4

v = 256.7 m/s

(c) The derivative of velocity function with respect to time gives the value of instantaneous acceleration.

So, a = dv / dt = 5.3 x 3 x 2 x t

a = 31.8 t

Put t = 4 s

a = 31.8 x 4 = 127.2 m/s^2

4 0

4 years ago

Professor Stefanovic is spinning a bucket of water by extending his arm and rotating his shoulder in class to show the effects o

anyanavicka [17]

Answer:

a ) 2.368 rad/s

b) 3.617 rad/s

Explanation:

the minimum angular velocity that Prof. Stefanovic needs to spin the bucket for the water not to fall out can be determined by applying force equation in a circular path

i.e

$F_{inward } = F_G + F_T$ ------ equation (1)

where;

$F_{inward} = m *a_c$

$F_{inward} = m*r* \omega^2$

Also

$F_G = m*g$

$F_T = 0$ since; that is the initial minimum angular velocity to keep the water in the bucket

Now; we can rewrite our equation as :

$mr \omega^2= mg + 0\\\omega^2 = \frac{m*g}{m*r}\\\omega^2 = \frac{g}{r}\\\omega = \sqrt{\frac{g}{r} \ \ } ------ equation \ \ \ {2}$

So; Given that:

The rope that is attached to the bucket is lm long and his arm is 75 cm long.

we have our radius r = 1 m + 75 cm

= ( 1 + 0.75 ) m

= 1.75 m

g = acceleration due to gravity = 9.81 m/s²

Replacing our values into equation (2) ; we have:

$\omega = \sqrt{ \frac{9.81}{1.75}}\\\omega = 2.368 \ rad/s$

b) if he detaches the rope and spins the bucket by holding it with his hand ; then the radius = 0.75 m

∴

$\omega = \sqrt{ \frac{9.81}{0.75}}\\\omega = 3.617 \ rad/s$

4 0

4 years ago

A dam is a structure built across a river to hold back the river's water. The flow of water through a dam is controlled by gates

Bess [88]

Potential is the first blank and kinetic is the second blank.

This should be correct

8 0

3 years ago

Read 2 more answers

A ray diagram without the produced image is shown.

4vir4ik [10]

A car acting as an object in front of a biconvex lens between F and 2 F on the object side of the lens. There is a light ray parallel to the principal axis that is bent through F on the image side of the lens. There is a ray straight through the center of the lens. The rays intersect below the x axis further than 2 F away from the lens and farther from the principal axis than the object is tall.

<u> The image produced by the lens is (b) inverted and real</u>

Explanation:

A real image occurs where the rays converge.

Real images can be produced both by the concave mirrors or converging lenses, but the condition is that the object of consideration is always placed far away from the mirror or the lens than the focal point, and thus the real image produced is inverted.

A car acting as an object in front of a biconvex lens between F and 2 F on the object side of the lens. There is a light ray parallel to the principal axis that is bent through F on the image side of the lens. There is a ray straight through the center of the lens. The rays intersect below the x axis further than 2 F away from the lens and farther from the principal axis than the object is tall.

<u> The image produced by the lens is (b) inverted and real</u>

8 0

3 years ago