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3 years ago

I load a 0.4 kg marble into a slingshot and shoot it directly upward by applying a 36.2 N force

Physics

1 answer:

kakasveta [241]3 years ago

8 0

Answer:50mph

Explanation:

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A concave mirror brings the sun's rays to a focus in front of the mirror. Suppose the mirror is submerged in a swimming pool but

Karo-lina-s [1.5K]

Answer:

The sun rays will be focused at the same distance from the mirror.

Explanation:

Laws of reflection do not depend on the refractive index n any form. Hence, the sun’s rays will be focused the same distance from the mirror.

8 0

3 years ago

A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 8.00î m/s. Find (a

sergejj [24]

a. The particle has position vector

$\vec r(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)t\right)\,\vec\imath+\left(\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\right)\,\vec\jmath$

$\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

b. Its velocity vector is equal to the derivative of its position vector:

$\vec v(t)=\vec r'(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath$

c. At $t=7.00\,\mathrm s$ , the particle has position

$\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath$

$\vec r(7.00\,\mathrm s)=\left(56.0\,\vec\imath+49.0\,\vec\jmath\right)\,\mathrm m$

That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of $\|\vec r(7.00\,\mathrm s)\|=\sqrt{(56.0\,\mathrm m)^2+(49.0\,\mathrm m)^2}=74.4\,\mathrm m$ away from the origin in a direction of $\theta=\tan^{-1}\dfrac{49.0\,\mathrm m}{56.0\,\mathrm m}=41.2^\circ$ relative to the positive $x$ axis.

d. The speed of the particle at $t=7.00\,\mathrm s$ is the magnitude of the velocity at this time:

$\vec v(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)\,\vec\jmath$

$\vec v(7.00\,\mathrm s)=\left(8.00\,\vec\imath+14.0\,\vec\jmath\right)\dfrac{\rm m}{\rm s}$

Then its speed at this time is

$\|\vec v(7.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+(14.0\dfrac{\rm m}{\rm s}\right)^2}=16.1\dfrac{\rm m}{\rm s}$

4 0

3 years ago

HELP⚠️⚠️

alexgriva [62]

Answer:

I think u are traeling at speed of light and not ur friend

Explanation:

4 0

3 years ago

Read 2 more answers

2 Jim keeps a can on the ramp. The can

IceJOKER [234]

Answer:

The description of the statement is summarized throughout the explanation section below.

Explanation:

The layer of the ramp must've been harsh, which exacerbated the toy movement to occur gradually. This same friction force was indeed starting to function on the remote control car as well as trying to stop that one to react quickly.
Jim does have to simplify the exterior including its ramp by scrubbing or greasing to overcome impact as well as make driving travel quicker.

6 0

3 years ago

A spaceprobe in outer space is flying with a constant speed of 1.530 km/s. The probe has a payload of 1363.0 kg and it carries 3

abruzzese [7]

Answer:

6.33 km/s

Explanation:

Given that :

A spaceprobe in outer space is flying with a constant speed $v_i$ = 1.530 km/s.