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Crazy boy [7]
3 years ago
6

What is the force due to gravity of a 38 kg student?

Physics
1 answer:
alukav5142 [94]3 years ago
6 0

Answer:

F_g = 372.78 N

Explanation:

Formula for force of gravity is;

F_g = mg

Where;

m is mass

g is acceleration due to gravity

We are given;

Mass = 38 kg

Acceleration due to gravity has a constant value of 9.81 m/s²

Thus;

F_g = 38 × 9.81

F_g = 372.78 N

You might be interested in
A 40kg load is raised to a height of 25m. If the operation requires 1 min, find the power required​
OlgaM077 [116]

Answer:

163.33 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 40 Kg

Height (h) = 25 m

Time (t) = 1 min

Power (P) =..?

Next, we shall determine the energy. This can be obtained as follow:

Mass (m) = 40 Kg

Height (h) = 25 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 40 × 9.8 × 255

E = 9800 J

Finally, we shall determine the power. This can be obtained as illustrated below:

Time (t) = 1 min = 60 s

Energy (E) = 9800 J

Power (P) =?

P = E/t

P = 9800 / 60

P = 163.33 Watts

Thus, the power required is 163.33 Watts

8 0
2 years ago
An object that is falling has the following type(s) of energy. Ignore air resistance.
Anton [14]
Potential and kinetic
6 0
2 years ago
Read 2 more answers
I need help for the first 3 plz.
madam [21]
1) hypothesis
2) data
3) method

I think these are correct.
7 0
3 years ago
Read 2 more answers
A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi
snow_lady [41]

Answer:

The velocity of the truck after this elastic collision is 15.7 m/s            

Explanation:

It is given that,

Mass of the car, m_1=1.1\times 10^3\ kg

Mass of the truck, m_2=2.3\times 10^3\ kg

Initial velocity of the car, u_1=22\ m/s

Initial velocity of the truck, u₂ = 0

After the collision the velocity of the car is, v₁ = -11 m/s

Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

1.1\times 10^3\times 22+0=1.1\times 10^3\times (11)+2.3\times 10^3\times v_2    

v_2=15.7\ m/s

So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).

4 0
3 years ago
A 0.11 N m torque is applied to a fan that was initially at rest. The fan has moment of inertia of 0.034 kg m2. Determine the ki
Mars2501 [29]

Answer:

2.72*10-3 Joules

Explanation:

From Newton's second law of motion

F=ma

\tau=I*\alpha

given

\tau= 0.11Nm\\\\I=0.034kgm^2\\\\t= 8s\\\\\alpha=?

\alpha =0.11/0.034\\\\\alpha=3.23 rad/s^2

the angular velocity is

\omega = \alpha/t\\\\\omega =3.23/8\\\\\omega =0.4 rad/s

KE= 1/2*I* \omega^2

KE=1/2*0.034*0.4^2\\\\KE=1/2*0.034*0.16\\\\KE=0.00272\\\\KE=2.72*10^-3J

4 0
2 years ago
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