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3 years ago

Please help me to solve this, I don’t really know what to do.

Physics

1 answer:

zalisa [80]3 years ago

5 0

Explanation:

Use Ohm's Law

V / R = I

I = 5A

V = 110V

R =?

Solve for R

if you need further help let me know

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What is Newton's law of motion

const2013 [10]

Explanation:

Newton’s first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force

2nd- is a quantitative description of the changes that a force can produce on the motion of a body.

3rd- states that when two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction.

3 0

3 years ago

Read 2 more answers

The wave length of violet light rounded to the nearest nanometer is a __ nm

nataly862011 [7]

Wavelength= speed / frequency

so.....3× 10^8 / 7.26×10^14
= .413× 10^(-6)

in scientific notation= 4.13×10^(-7)

in nanometer = 413 nm

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3 years ago

A man with a mass of 60 kg rides a bike with a mass of 13 kg. What is the force needed to accelerate the bike at 0.90 m/s2?

Scrat [10]

The total mass of the things on the road is 60 + 13, which is 73kg.

mass • acceleration = F

73 x 0.9m/s^2 = 65.7N

F = 65.7 N

7 0

2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

4 years ago

Whats the word for when "velocity equals 0 and direction changes" Its 14 letters, and I have _ _ _ _ i _ _ _ _ _ h_ i _ _ t

alisha [4.7K]

Answer:

Maximum Height

Explanation:

The maximum height is the highest point reached by a projected body. At this point, final velocity, v of the body is equal to 0; because upward motion is at its peak and the body start to fall again.

Hence, final velocity v is always 0 at this maximum point and the direction of motion changes from upward to downward.

Final velocity at this maximum height is zero because of the directional change experied by the object and the fact that upward motion of the body terminates at this point.

6 0

4 years ago