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patriot [66]
2 years ago
7

A 1400 kg car traveling in the positive direction takes 11.0 seconds to slow from 25.0 meters per second to 12.0 meters per seco

nd. What is the average force on the car during this time?
Physics
1 answer:
Alex777 [14]2 years ago
6 0

Answer:

–1652 N. The negative sign indicate that the force is in opposite direction to car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 1400 kg

Time (t) = 11 s

Initial velocity (u) = 25 m/s

Final velocity (v) = 12 m/s

Force (F) applied on the car =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Time (t) = 11 s

Initial velocity (u) = 25 m/s

Final velocity (v) = 12 m/s

Acceleration (a) =?

a = (v – u) /t

a = (12 – 25) /11

a = –13/11

a = –1.18 m/s²

Finally, we shall determine the force on the car as follow:

Acceleration (a) = –1.18 m/s²

Mass (m) of car = 1400 kg

Force (F) applied on the car =?

F = ma

F = 1400 × –1.18

F = –1652 N

Thus, the force on the car is –1652 N. The negative sign indicate that the force is in opposite direction to car.

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Your outlets at home are rated at 120 V, i.e. the two prongs have on average a potential difference of 120V. If you transfer 2.7
hodyreva [135]

Answer:

E =230.4 MJ

Explanation:

As 1 mole of electron =  6X 10^23 particles.

charge of an electron is 1.6 X 10 ^-19 C

Finding Charge:

(6X10^23 ) (2.7)(1.6X10^-19 C)

i.e. 192 K C

now  to find the energy released from electrons

V=E/q

E=V X q

i.e E = 120 V X 192 K C

E =230.4 MJ

4 0
3 years ago
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Tamiku [17]
True, they represent the direction of motion
8 0
3 years ago
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn
KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

6 0
3 years ago
A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
castortr0y [4]

The vertical component of force exerted by the hi.nge on the beam will be,142.10N.

To find the answer, we need to know more about the tension.

<h3>How to find the vertical component of the force exerted by the hi.nge on the beam?</h3>
  • Let's draw the free body diagram of the system.
  • To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.

                      F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\

  • To find the answer, we have to find the tension,

                     Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N

  • Thus, the vertical component of the force exerted by the hi.nge on the beam will be,

                F_V=(29*9.8)-(169.43*sin57)=142.10N

Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.

Learn more about the tension here:

brainly.com/question/28106868

#SPJ1

5 0
2 years ago
Read 2 more answers
Convert 4 radians to degrees A. 242.6° B. 9.6° C. 229.2° D. 6.8°
muminat
The central angle of a circle is 360° or 2π radians.

Therefore
1 radian = (360 degrees)/(2π radians) = 180/π degrees/radian.
4 radians = (4 radians)*(180/π degrees/radian) = 229.18 degrees.

Answer: C.  229.2°
8 0
3 years ago
Read 2 more answers
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