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Reil [10]
3 years ago
11

A 2100 kg car starts from rest and accelerates at a rate of 2.6 m/s2 for 4.0 s. Assume that the force acting to accelerate the c

ar is acting in the same direction as its motion. How much work has the car done?
Physics
1 answer:
Reika [66]3 years ago
8 0
When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.

a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s

Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ
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If a body travels half its total path in the last 1.50 s of its fall from rest, find the total time of its fall (in seconds).
svetoff [14.1K]

Answer:

time to fall is 3.914 seconds

Explanation:

given data

half distance time = 1.50 s

to find out

find the total time of its fall

solution

we consider here s is total distance

so equation of motion for distance

s = ut + 0.5 × at²   .........1

here s is distance and u is initial speed that is 0 and a is acceleration due to gravity = 9.8 and t is time

so for last 1.5 sec distance is 0.5 of its distance so equation will be

0.5 s = 0 + 0.5 × (9.8) × ( t - 1.5)²     ........................1

and

velocity will be

v = u + at

so velocity v = 0+ 9.8(t-1.5)    ..................2

so first we find time

0.5 × (9.8) × ( t - 1.5)² = 9.8(t-1.5)  + 0.5 ( 9.8)

solve and we get t

t = 3.37 s

so time to fall is 3.914 seconds

6 0
4 years ago
A converging lens has a focal length of 20 cm. An object 1 cm tall is placed 10 cm from the center of the lens. What is the heig
SCORPION-xisa [38]

Answer: 2 cm

Explanation:

Given , for a converging lens

Focal length : f=20\ cm

Height of object : h=1\ cm

Object distabce from lens : u=-10\ cm

Using lens formula: \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}, we get

\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}, where v = image distance from the lens.

On solving aboive equation , we get

\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm

Formula of Magnification : m=\dfrac{v}{u}=\dfrac{h'}{h} , where h' is the height of image.

Put value of u, v and h in it , we get

\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm

Hence, the height of the image is 2 cm.

3 0
3 years ago
If a farsighted person has a near point that is 0.600 mm from the eye, what is the focal length f2f2f_2 of the contact lenses th
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Answer:

0.22mm

Explanation:

A far sighted person is a person suffering from long sightedness i.e such individual can only see far distant object clearly but not near distant object. The defect is corrected using convex lens.

Since convex lens is used, the focal (f) length of the lens is positive and the image distance (v) is also positive.

Using the lens formula,

1/f = 1/u + 1/v

Where u is the object distance = 0.35mm

v = 0.6mm

1/f = 1/0.35+1/0.6

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f = 1/4.53

f = 0.22mm

The focal length of the contact lenses will be 0.22mm

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3 years ago
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Answer:

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Electromagnetic waves are transverse waves composed by the perpendicular oscillating electric and magnetic fields.

EM waves have both Electrical and magnetic features.

they travel in the velocity of light (3*10⁸ ms⁻¹)

they does not require any media to travel. It has two perpendicular electric field and the magnetic field which are perpendicular to each other

They travel perpendicular to each of those electric and magnetic fields.  

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Answer:

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Explanation:

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