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weqwewe [10]
3 years ago
7

A car whose total mass is 800kg travelling with a uniform velocity of 20m/s suddenly observes a stationary dog 50m ahead on its

path. If the total force applied on the braking system on the car by the driver is 2000N, what would happen? ​
Physics
1 answer:
maxonik [38]3 years ago
7 0

Answer:

The driver hits the stationery dog because the applied force is less than required force

Explanation:

Kinetic energy will be given by

KE=0.5mv^{2} where m is the mass of the vehicle and v is the speed/velocity of the vehicle.

Substituting 800 Kg for m and 20 m/s for v we obtain

KE=0.5*800*(20 m/s)^{2}=160,000

Frictional force by vehicle pads is given by

Fr=\frac {KE}{d} where d is the distance moved

Substituting 160000 for KE and 50 m for d we obtain

Fr=\frac {160000}{50}=3200 N

Therefore, the vehicle hits the dog since the required force is 3200N but the driver applied only 2000 N

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A rock of mass m is thrown straight up into the air with initial speed |v0 | and initial position y = 0 and it rises up to a max
REY [17]

Answer:

Explanation:

Case 1:

mass = m

initial velocity = vo

final velocity = 0

height = y

Use third equation of motion

v² = u² - 2as

0 = vo² - 2 g y

y = vo² / 2g       ... (1)

Case 2:

mass = 2m

initial velocity = 2vo

final velocity = 0

height = y '

Use third equation of motion

v² = u² - 2as

0 = 4vo² - 2 g y'

y ' = 4vo² / 2g

y' = 4 y

Thus, the second rock reaches the 4 times the distance traveled by the first rock.

7 0
3 years ago
A 75-kg piano is hoisted on a crane and delivered throughout the window of a 6th story apartment (20 meters above ground). What
ipn [44]

Answer:

P = 14700 J

Explanation:

Given that,

Mass of a piano, m = 75 kg

It is delivered throughout the window of a 6th story apartment which is 20 m above the ground.

We need to find the potential energy of the piano. It is given by :

P = mgh

Putting all the values,

P = 75 kg × 9.8 m/s² × 20 m

P = 14700 J

So, the potential energy of the piano is 14700 J.

7 0
3 years ago
An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c
Aleks04 [339]

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

                      \frac{1}{f}=\frac{1}{v}-\frac{1}{u}

                      \frac{1}{10}=\frac{1}{v}-\frac{1}{-5}

                      \frac{1}{v}=\frac{1}{10}-\frac{1}{5}

                      v = -10 cm

The magnification factor of lens

                     m=\frac{-10}{-5}

                     m = 2

                     m=\frac{h'}{h}

                     h'=h\times m

                     h'=3\times 2

                     h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

4 0
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What is the wavelength of a wave that has a speed of 350 meters/second and a frequency of 140 hertz?In meters
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Apply:
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it is B

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