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brilliants [131]
3 years ago
11

A skydiver has a mass of 65 kg. Calculate the weight of the skydiver.

Physics
1 answer:
Vinil7 [7]3 years ago
3 0
F=m*a
F=65 kg *9.8 m/s^2
F=637 N (Newtons) — this is the weight
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An 12 N force is applied to a 1 kg object. What is the magnitude of the objects acceleration?
Sauron [17]

Answer:

a=12 m/s²

Explanation:

Newton's second law of motion states that the acceleration of a body is directly proportional to the force applied and takes place in the direction of force.

This can be summarized as: F=ma, where m is the mass of the object on which force F acts. a is the acceleration due to the force applied.

12N= 1kg×a

a=12N/1kg

a=12m/s²

6 0
3 years ago
An 800 kHz radio signal is detected at a point 3.2 km distant from a transmitter tower. The electric field amplitude of the sign
hammer [34]

Answer:

1.07 nT

Explanation:

We know that E/B = c where E = electric field amplitude = 320 mV/m = 0.32 V/m, B = magnetic field amplitude and c = speed of light = 3 × 10⁸ m/s.

So, B = E/c

Substituting E and c into B, we have

B = E/c

= 0.32 V/m ÷ 3 × 10⁸ m/s

= 0.1067 × 10⁻⁸ T

= 1.067 × 10⁻⁹ T

= 1.067 nT

≅ 1.07 nT

3 0
3 years ago
How is temperature related to kinetic energy
stiks02 [169]
Temperature is the average kinetic energy of an object
7 0
3 years ago
A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozz
gladu [14]

This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle?  Assume the specific heat ratio is 1.2.

Answer:

- stagnation pressure is 274.993 Mpa

- the stagnation temperature Tt is 4048 K

Explanation:

Given the data in the question;

To determine the stagnation pressure and temperature in the exit plane of the nozzle;

we us the expression;

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

where Pt is stagnant pressure = ?

P is static pressure = 4 MPa = 4 × 10⁶ Pa  

Tt is stagnation temperature = ?

T is the static temperature  = 2000 K

γ is ratio of specific heats = 1.2

M is Mach number M = 3.2

we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = P(1 + (γ-1 / 2) M²)^(γ/γ -1)

Pt = 4 × 10⁶(1 + (1.2-1 / 2) 3.2²)^(1.2/1.2 -1)

Pt = 4 × 10⁶ × 68.7484

Pt = 274.993 × 10⁶ Pa

Pt = 274.993 Mpa

Therefore stagnation pressure is 274.993 Mpa

Now, to get our stagnation Temperature

Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa =  ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 =  Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

6 0
3 years ago
Another athlete using a different spring exerts an average force of 400 N to enable her
babymother [125]

Answer:84Nm

Explanation:

force=400N

Distance=0.210m

Workdone=force x distance

Workdone=400 x 0.210

Workdone=84Nm

6 0
3 years ago
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