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Bezzdna [24]
3 years ago
14

Help meeeee!!! ⇓⇓⇓⇓⇓

Physics
2 answers:
Dafna11 [192]3 years ago
6 0

I believe you are correct, it is B: Diagnostic Services.

Diagnostic services are services like the staff at hospitals and the people who run machines that are related to medical needs.

<em>If this is incorrect, please, don't refrain to tell me.</em>

Rudiy273 years ago
5 0
Yea I agree I think it is B
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A ball is dropped off a very tall canyon ledge.
oksano4ka [1.4K]

Answer:

V = 49.05 [m/s]

Explanation:

We can easily find the result using kinematics equations, first, we will find the distance traveled during the 5 seconds.

y =y_{o}+(v_{o}*t)+(\frac{1}{2}*g*t^{2} )

where:

Yo = initial position = 0

y = final position [m]

Vo = initial velocity = 0

t = time = 5 [s]

g = gravity aceleration = 9.81 [m/s^2]

The initial speed is zero, as the body drops without imparting an initial speed. Therefore:

y = 0 + (0*5) + (0.5*9.81*5^2)

y = 122.625[m]

Now using the following equation we can find the speed it reaches during the 5 seconds.

v_{f} ^{2}= v_{i} ^{2}+(2*g*y)\\v_{f}=\sqrt{2*9.81*122.625} \\v_{f}=49.05 [m/s]

6 0
4 years ago
You start at (8, 1). You move right 2 units. Where do you end?
ivann1987 [24]

I'm pretty sure should be shifting your transformation those two units to the right from the original function.

Moving right = subtracting

Moving left = adding

4 0
3 years ago
A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0
4 years ago
A constant force of 5KN pulls a crate along a distance of 15 m in 75s.What is the power​
xxMikexx [17]

Explanation:

We know,

1KN = 1000N

Then, Force(F) = 5*1000N

=5000N

Here,

Power (P)=Work(W)/Time(T)

=Force * distance/ Time (W = F*s)

= 5000*15/75

=1000

So, The power of body or object is 1000Watt.

I hope this will be helpful for you.

8 0
4 years ago
Can u help me with the answer
Mice21 [21]
C. 6.7 hope it helps:))
7 0
3 years ago
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