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Charra [1.4K]
3 years ago
14

A robot standing on a cliff shoots a ball upwards with an initial speed of 30 m/s. What is the height of the cliff given that th

e ball reaches the bottom of the cliff 8 s after the shoot? (Take g = 10 m/s^2 and the height of the robot is negligible.)
A 25 m
B 45 m
C 80 m
D 145 m​

Physics
1 answer:
valina [46]3 years ago
5 0

Answer:

C 80 m

Explanation:

Given:

v₀ = 30 m/s

a = -10 m/s²

t = 8 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²

Δy = -80 m

The ball lands 80 m below where it started.  So the height of the cliff is 80 m.

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An object of mass 0.40 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into an up-and-down simple harmonic m
Sergeeva-Olga [200]

Answer:

a = 2 m/s2

Explanation:

we know from newtons 2nd law

F = ma.

we also know that from hookes law we have

F = kx

equate both value of force to get value of acceleration

kx = ma,

where,

k is spring constant = 8.0 N/m

x is maximum displacement  0.10 m

m is mass of object 0.40 kg

a = \frac{kx}{m}

     = \frac{8 *0 .10}{0.40}

a = 2 m/s2

5 0
3 years ago
If a 100-N net force acts on a 50-kg car, what will the acceleration of the car be?
Tema [17]
Newton's 2nd law of motion: 

                             Force = (mass) x (acceleration)

Divide each side by (mass):     

                             Acceleration = (force) / (mass)

                                               =  (100 N) / (50 kg)

                                               =  2 m/s²  


5 0
3 years ago
I need help on the data section of the circuit design lab on Edg.
Arte-miy333 [17]

I hope it's not too late, but here you go

8 0
3 years ago
The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa
Sonja [21]

Answer:

m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

c_{pA} > c_{pB}

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

T_{iA}= T_{iB}

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

Q = m c_p \Delta T

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

Q_A = Q_B

And if we replace the formula for sensible heat we got:

m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B

And if we replace for the change of the temperature we got:

m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})

And since T_{iA}= T_{iB}= T we have this:

m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

4 0
3 years ago
A 9.0-kg bowling ball on a horizontal, frictionless surface experiences a net force of 6.0 n. what will be its acceleration?
Vladimir [108]

This question involves the concepts of Newton's Second Law of Motion.

The acceleration of the bowling ball will be "0.67 m/s²".

<h3>Newton's Second Law of Motion</h3>

According to Newton's Second Law of Motion, when an unbalanced force is applied on an object, it produces an acceleration in it, in the direction of the applied force. This acceleration is directly proportional to the force applied and inversely proportional to the mass of the object. Mathematically,

F=ma\\\\a=\frac{F}{m}

where,

  • a = acceleration = ?
  • F = Magnitude of the applied force = 6 N
  • m = Mass of the ball = 9 kg

Therefore,

a=\frac{6\ N}{9\ kg}

a = 0.67 m/s²

Learn more about Newton's Second Law of Motion here:

brainly.com/question/13447525

#SPJ1

7 0
2 years ago
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