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NeX [460]
2 years ago
6

Q12. How much force is applied on the body when 150 joule of work

Physics
1 answer:
ra1l [238]2 years ago
4 0

Force = Work/distance

Force = 150/10

          =  15 Newtons

Force = 15 Newtons

Therefore, 15 newtons of force is applied to the body when 150 joules of work

is done in displacing the body through a distance of 10m in the  direction of the force.

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The ratio of output power to input power, in percent, is called?
Harrizon [31]
That's efficiency. There's no law that it must be stated in percent.
4 0
2 years ago
Read 2 more answers
Please help Need good grade
ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the  range

Using formula of range

R=\dfrac{v^2\sin(2\theta)}{g}...(I)

The range for the other angle is

R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}....(II)

Here, distance and speed are same

On comparing both range

\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}

\sin(2\theta)=\sin(2\times(\alpha-\theta))

\sin120=\sin2(\alpha-60)

120=2\alpha-120

\alpha=\dfrac{120+120}{2}

\alpha=120^{\circ}

Hence, The other angle is 120°

6 0
3 years ago
Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.
mr_godi [17]

Answer:

The magintude of the acceleration for both objects is 3.11m/s^2

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

T_A-m_A*g=m_A*a

and for mB (descending):

T_B-m_B*g=-m_B*a

T_A=T_B

because the two boxes has the same acceleration because they are attached together:

m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2

So the magintude of the acceleration for both objects is 3.11m/s^2

5 0
3 years ago
HELPPP PLEASEEE!!!!!
devlian [24]

Answer:

1) The speed of sound increases

2)  440 Hz

3)  29°C

4)  17°C

5) 434 Hz

6)  12 m/s

7)  17.3 m

Explanation:

1) The speed of sound increases

2) V = f×λ

f = V/λ = 343/0.78 = 439.744 ≈ 440 Hz

3) V = f×λ

512 × 0.68 = 348.16 m/s

348.16 - 331 = 17.16

T = 17.16/0.6 = 28.6 ≈ 29°C

4) Increase in speed = 350 - 340 = 10

Increase in temperature = 10/0.6 = 16.67° ≈ 17°C

5) f = V/λ = 343/0.79 = 434 Hz

6) 331 + 0.6×30 - (331 × 0.6 ×10) = 12 m/s

7) V = 331 + 0.6×25 = 346m/s

λ = 346/20 = 17.3 m

5 0
3 years ago
Question 2 (1 point)
Tju [1.3M]

Answer:

I know someone anwsered but it would be 400M

Explanation:

i initial velocity (u)=10m/s

acceleration (a)=0

time taken (t) =40s

then distance (s)=u t +1/2 a t^2

s= u t +0 (as a is 0)

s= 10 x 40

s= 400M

7 0
3 years ago
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