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san4es73 [151]
1 year ago
13

According to the work-energy theorem, if work is done on an object, its potential and/or kinetic energy changes. Consider a car

that accelerates from rest on a flat road. What force did the work that increased the car’s kinetic energy?
1. the force of the car engine

2. air resistance

3. the friction between the road and the tires

4. gravity
Physics
1 answer:
Leni [432]1 year ago
6 0

Answer:

The force of the car engine.

Explanation:

The work- energy theorem states that the work done on an object is equal to the change in its kinetic energy. Its expression is given by :

W=\dfrac{1}{2}m(v^2-u^2)

Also, W = F.d

Fd=\dfrac{1}{2}m(v^2-u^2)

Where

F is the force applied by the engine of car

d is the displacement

m is the mass of an object

u is the initial speed

v is the final speed

So, the force of the car engine increased the car’s kinetic energy. Hence, this is the required solution.

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zavuch27 [327]

Answer:

The value of R is 1.72\times10^{11}\ m.

(B) is correct option.

Explanation:

Given that,

In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

We need to calculate this for value of R

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

R=1.89\times10^{11}\ m

So, The nearest option of the value of R is 1.72\times10^{11}\ m

Hence, The value of R is 1.72\times10^{11}\ m.

6 0
2 years ago
Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear
Sidana [21]

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

Sm = √(2.51×10^-4 / 5.66×10^8)

Sm = √0.443×10^-12

Sm = 6.65×10^-13 m.

8 0
2 years ago
At which lunar phase(s are tides least pronounced (e.g., the lowest high tides? at which lunar phase(s are tides least pronounce
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Both first and third quarters. (:
7 0
2 years ago
A large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is subme
Lerok [7]

The density of the substance is<u> 10.5 g/cm³.</u>

The jewelry is made out of <u>Silver.</u>

Density ρ is defined as the ratio of mass <em>m</em> of the substance to its volume V<em>. </em> The cylinder contains a volume <em>V₁ of water</em> and when the jewelry is immersed in it, the total volume of water and the jewelry is found to be V₂.

The volume <em>V</em> of the jewelry is given by,

V=V_2 -V_1

Substitute 48.6 ml for <em>V₁ </em>and 61.2 ml for V₂.

V=V_2 -V_1\\ =61.2 ml -48.6 ml\\ =12.6 ml

calculate the density ρ of the jewelry using the expression,

\rho =\frac{m}{V}

Substitute 132.6 g for <em>m</em> and 12.6 ml for <em>V</em>.

\rho =\frac{m}{V}\\ =\frac{132.6 g}{12.6 ml} \\ =10.5 g/ml

Since 1 ml=1 cm^3,

The density of the jewelry is <u> 10.5 g/cm³.</u>

From standard tables, it can be seen that the substance used to make the jewelry is <u>silver</u><em><u>, </u></em>which has a density 10.5 g/cm³.



4 0
2 years ago
A 60 kilogram astronaut weighs 96 newtons on the surface of the moon. calculate the acceleration due to gravity on the moon.
34kurt
Given: Mass m = 60 Kg

           Weight  W = 96 N

Required: Acceleration due to gravity, g = ?

Formula:  W = mg

                g = W/g

                g = 96 Kg.m/s²/60 Kg   (note: this is the derive unit for Newton "N")

                g = 1.6 m/s²

6 0
1 year ago
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