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vova2212 [387]
3 years ago
12

What is the minimum force required to increase the energy of a car by 69 J over a distance of 42 m? Assume the force is constant

.
Group of answer choices

0.61 N

69 N

2900 N

1.6 N

110 N
Physics
1 answer:
sineoko [7]3 years ago
3 0
ANSWER: 1.6N (forth option)



EXPLANATION:
E= Force x displacement
69=F(42)
69/42 = F
F=1.64 N

Answer : 1.6N
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An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance
jeka94

Answer:

1.8\times 105 N/C

Explanation:

We are given that

u=2\times 10^7 m/s

v=4\times 10^7 m/s

d=1.9 cm=\frac{1.9}{100}=0.019 m

Using 1m=100 cm

We have to find the electric field strength.

v^2-u^2=2as

Using the formula

(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)

16\times 10^{14}-4\times 10^{14}=0.038a

0.038a=12\times 10^{14}

a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2

q=1.6\times 10^{-19} C

Mass of electron,m=9.1\times 10^{-31} kg

E=\frac{ma}{q}

Substitute the values

E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}

E=1.8\times 105 N/C

7 0
4 years ago
Why Study Science? Help me out
Mashutka [201]
All but the last one!
6 0
1 year ago
Read 2 more answers
A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i
tresset_1 [31]

Answer:

a) Acceleration of the car is given as

a_{car} = -21 m/s^2

b) Acceleration of the truck is given as

a_{truck} = 10.15 m/s^2

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

m_1v_1 + m_2v_2 = (m_1 + m_2)v

now we have

1400 (6.32) + 2900(0) = (1400 + 2900) v

v = 2.06 m/s

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

a_{car} = \frac{v_f - v_i}{t}

a_{car} = \frac{2.06 - 6.32}{0.203}

a_{car} = -21 m/s^2

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

a_{truck} = \frac{v_f - v_i}{t}

a_{truck} = \frac{2.06 - 0}{0.203}

a_{truck} = 10.15 m/s^2

5 0
3 years ago
horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
3 years ago
Read 2 more answers
For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equat
andrezito [222]

Answer:

Radius of the solenoid is 0.93 meters.

Explanation:

It is given that,

The magnetic field strength within the solenoid is given by the equation,

B(t)=5t\ T, t is time in seconds

\dfrac{dB}{dt}=5\ T

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m

The electric field due to changing magnetic field is given by :

E(2\pi x)=\dfrac{d\phi}{dt}

x is the distance from the axis of the solenoid

E(2\pi x)=\pi r^2\dfrac{dB}{dt}, r is the radius of the solenoid

r^2=\dfrac{2xE}{(dE/dt)}

r^2=\dfrac{2\times 2\times 1.1}{(5)}

r = 0.93 meters

So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.

4 0
4 years ago
Read 2 more answers
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