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3 years ago

If m=120kg and a=15m/s2, what is the force

Physics

1 answer:

Scorpion4ik [409]3 years ago

6 0

Answer:

F= 1800N

Explanation:

the equation for force is F= ma

so plug in the numbers: F= (120)(15)

solve this to get F= 1800N

tip: don't forget to add the units when writing your answer :)

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A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c

mote1985 [20]

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.

F ∝ e

F = ke

where;

k is the spring constant

k = F / e

k = 40 / 0.04

k = 1000 N/m

Therefore, the value of the spring constant of this spring is 1000 N/m

7 0

4 years ago

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau

suter [353]

Answer:

actual efficiency is 47.78 %

Carnot efficiency is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy = 7.0 × 10^9 cal (4.184 J/1 cal) = 29.29 × 10^9 J

actual efficiency = 1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency = 0.4778

actual efficiency is 47.78 %

and

Carnot efficiency is = 1 - ( 703 / 2173 )

so Carnot efficiency is = 0.67648

Carnot efficiency is 67.65 %

and

power output = work / time

power output = 1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0

3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

sdas [7]

Answer:

0.34 m

Explanation:

From the question,

v = λf................ Equation 1

Where v = speed of sound, f = frequency, λ = Wave length

Make λ the subject of the equation

λ = v/f............... Equation 2

Given: v = 340 m/s, f = 500 Hz.

Substitute these values into equation 2

λ = 340/500

λ = 0.68 m

But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length

Therefore,

λ/2 = 0.68/2

λ/2 = 0.34 m

Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m

6 0

3 years ago

PLZ HELP IM REALLY CONFUSED a boy is playing catch with his friend. He throws the ball straight up. When it leaves his hand, the

erastovalidia [21]

Answer:

K.E = 100 J

Final P.E = 100 J

Explanation:

The kinetic energy of any object can be given by the following formula:

$K.E = (\frac{1}{2})mv^{2}$

where,

K.E = Kinetic Energy

m = mass of ball = 2 kg

v = speed of ball

Initially, v = 10 m/s. Therefore, the initial K.E is given as:

$K.E = (\frac{1}{2})(2\ kg)(10\ m/s)^{2}$

Now, at the highest point the K.E of the ball becomes zero. because the ball stops for a moment at the highest point and its velocity becomes zero. So, from Law of Conservation of energy:

Initial K.E + Initial P.E = Final K.E + Final P.E

Initial P.E is also zero due to zero height initially.

K.E + 0 = 0 + Final P.E

<u>Final P.E = 100 J</u>

3 0

3 years ago

The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, dete

sweet [91]

Answer:

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

$F_n = 1.72 \times 10^4$

Explanation:

Since we need to cross both the loops so least speed at the bottom must be

$v = \sqrt{5 R g}$

also by energy conservation this is gained by initial potential energy

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

so we will have

$\sqrt{2gh} = \sqrt{5Rg}$

now we have

$h = \frac{5R}{2}$

here we have

R = 7.5 m

so we have

$h = \frac{5(7.5)}{2}$

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

now when it reach point C then the speed will be

$mgh - mg(2R_c) = \frac{1}{2]mv_c^2$

$v_c^2 = 2g(h - 2R_c)$

$v_c = 13.1 m/s$

now normal force at point C is given as

$F_n = \frac{mv_c^2}{R_c} - mg$

$F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)$

$F_n = 1.72 \times 10^4$

7 0

4 years ago