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bonufazy [111]
3 years ago
8

An object of size 5cm is placed in front of a concave mirror of focal length of magnitude 20 cm at a distance of 40cm in front o

f the mirror. The image formed is
Physics
1 answer:
zheka24 [161]3 years ago
7 0

Are there any options?

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How to find acceleration?
Yanka [14]
Acceleration = Change in Velocity / time
 
                a  =  (v - u) / t

Where v = final velocity in m/s              
            u = initial velocity in m/s 
            t = time in seconds.
            a = acceleration in m/s²

A proper record of the changes in velocity with the corresponding time would help find the acceleration.  
4 0
2 years ago
Read 2 more answers
A 1500 kg car, initially traveling at 22.0 m/s, hits its brakes and skids to a stop. Determine the work done by friction.
Sedbober [7]

Answer:

<em>The work done by the car is 363 kJ</em>

Explanation:

Work : Work is said to be done when a Force moves an object through a certain distance. Work and Energy are interchangeable because they have the same unit. The unit of work is Joules (J).

Mathematically work done can be expressed as,

E = W = 1/2mv²

W =  1/2mv²................................ Equation 1

Where E = Energy, W = work done, m = mass of the car, v = velocity of the car

<em>Given: m=1500 kg, v=22 m/s</em>

<em>Substituting these values into equation 1</em>

<em>W = 1/2(1500)(22)²</em>

<em>W = 750 × 484</em>

<em>W = 363000 J</em>

<em>W = 363 kJ</em>

<em>Thus the work done by the car is 363 kJ</em>

8 0
2 years ago
A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg   , acting in upward direction

T - mg = m a

T =  mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D )  If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

= 63 x ( 9.81 - .6 )

= 580.23 N

6 0
3 years ago
A college friend of yours who has been postponing taking any science courses hears you talking about the generation of nuclear e
mihalych1998 [28]

Answer:

It's held together by the nuclear force.

Explanation:

There are <em>more</em> elemental forces than just the electromagnetic one. In this case, it is the nuclear force (called also strong force) the one that holds the nucleus together because it is stronger than the electromagnetic force over such short distances as the one inside the atomic nucleus.

8 0
3 years ago
What is the lift (in newtons) due to Bernoulli's principle on a wing of area 76 m2 if the air passes over the top and bottom sur
AveGali [126]

Answer:

So lift will be 30.19632 N

Explanation:

We have given area of the wing a=76m^2

We know that density of air d=1.29kg/m^3

Speed at top surface v_2=290m/sec and speed at bottom surface v_1=150m/sec

According to Bernoulli's principle force is given by

F=A\times d\times \frac{v_2^2-v_1^2}{2}=76\times 1.29\times \frac{290^2-150^2}{2}=3019632N

4 0
3 years ago
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