All categories

3 years ago

PLEASE HELP! LAST DAY TO TURN IN AND IM SO BEHIND!!!

Physics

2 answers:

kolbaska11 [484]3 years ago

7 0

<h3>Question 1</h3>

Answer

option C) velocity

Explanation

acceleration = Δv ÷ Δt

<h3>Question 2</h3>

Answer

option C) m/s²

Explanation

Δv ÷ Δt

= m/s ÷ s

= m/s x 1/s

= m/s²

<h3>Question 3</h3>

Answer

option B) velocity has both direction and speed.

That is why velocity can be negative but speed can not and velocity is rate of change of displacement where as speed is rate of change of distance.

slamgirl [31]3 years ago

6 0

Velocity, m/s², and velocity has both speed and direction.

You might be interested in

What is homeostasis?

SashulF [63]

<span>Homeostasis maintains internal balance of the cell.-- Things like shiver if cold sweating if hot to maintain internally </span>

3 0

4 years ago

Read 2 more answers

Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black

arlik [135]

Answer:

$833.4801043*10^6N$ on ear that is closer to Black hole.

$13.83803929*10^6N$ On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

$F =\frac{m_{1}*m_{2} }{r^2} G$

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

r = $120km-\frac{6}{100000} km=119.99994km=119999.94m$

Note, we have subtracted because ear is closer to black hole.

plugging all this in formula gives.

$F = 833.4801043*10^6N$

That is tension of ear.

(2) Force on ear that is further from black hole.

$F =\frac{m_{1}*m_{2} }{r^2} G$

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

this time r is further away from black hole so it would be.

$r = 120km+\frac{6}{100000} km = 120000.06m$

Plugging this all in we get

$F = 13.83803929N$

and that is tension on ear that is further from black hole.

Notice the tension difference, and order of magnitude of tension,it is enormous .

this astronaut is lethally close to black hole.

5 0

3 years ago

A military helicopter on a training mission is flying horizontally at a speed of 90.0 m/s when it accidentally drops a bomb (for

Elena-2011 [213]

Answer:

1) 10.1 s 2) 909 m 3) 90.0 m/s 4) -99m/s 5) just over the bomb.

Explanation:

In the vertical direction, as the bomb is dropped, its initial velocity is 0.
So, we can find the time required for the bomb to reach the earth, applying the following kinematic equation for displacement:

$\Delta y = \frac{1}{2}*a*t^{2} (1)$

where Δy = -500 m (taking the upward direction as positive).
a=-g=-9.8 m/s²
Replacing these values in (1), and solving for t, we have:

$t =\sqrt{\frac{2*\Delta y}{-g}} = \sqrt{\frac{2*(-500m)}{-9.8m/s2}} = 10.1 s$

The time required for the bomb to reach the earth is 10.1 s.

In the horizontal direction, once released from the helicopter, no external influence acts on the bomb, so it will continue moving forward at the same speed. that it had, equal to the helicopter.
As the time must be the same for both movements, we can find the horizontal displacement just as the product of this speed times the time, as follows:

$x = v_{0x} * t = 90.0 m/s * 10.1 s = 909 m.$

The horizontal component of the bomb's velocity is the same that it had when left the helicopter. i.e. 90 m/s.

In order to find the vertical component of the bomb's velocity just before it strikes the earth, we can apply the definition of acceleration, remembering that v₀ = 0, as follows:

$v_{f} = -g*t = -9.8 m/s2*10.1 s = -99 m/s$

If the helicopter keeps flying horizontally at the same speed, it will be always over the bomb, as both travel horizontally at the same speed.
So, when the bomb hits the ground, the helicopter will be exactly over it.

8 0

4 years ago

Above questions which on paper

Luden [163]

Answer:

Explanation:

hghgghfgfgc

6 0

3 years ago

The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this in powers of ten i

Levart [38]

It could be express as (A) 14 x 10 to the tenth power or (b) 4.415 x 10 to the seventeenth power.

8 0

4 years ago