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tatuchka [14]
3 years ago
12

A metal box, attached to a small parachute, is dropped from a helicopter. explain in terms of the forces acting; why:

Physics
2 answers:
faust18 [17]3 years ago
6 0

Answer:

1

Explanation:

Anna35 [415]3 years ago
3 0
(i) because there's gravity force pulling it downwards. 

(ii) because the force on the metal box is gone (either it has landed on the ground or something else happened) that means that the metal box is no longer moving but it has constant velocity/uniform velocity. 

I'm sorry if it's wrong but that's how I see it 
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The radius of Saturn is about 10 times the radius of Venus and the mass is about 100 times that of Venus. How much larger is the
lora16 [44]

let the mass of Venus is M then mass of Saturn is 100 M

similarly if the radius of Venus is R then the radius of Saturn is 10 R

now the force of gravity on a man of mass "m" at the surface of Venus is given by

F_1 = \frac{GMm}{R^2}

now similarly the gravitational force on the man if he is at the surface of Saturn

F_2 = \frac{G*100M*m}{(10R)^2}

F_2 = \frac{GMm}{R^2}

so here if we divide the two forces

\frac{F_1}{F_2} = 1

so here we can say

F1 = F2

so on both planets the gravitational force will be same

7 0
3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
The sunspot cycle is a pattern of solar activity where the average number of sunspots gradually _______________________________
zzz [600]

Answer:

option D

Explanation:

Sunspots are the spot that appears on the sun, this spot appears darker than the surrounding surface of the sun.

Sun magnetic field goes through a cycle and this cycle is called the Sunspot cycle. Every 11 years the magnetic field of the sun completely flips. This sunspot cycle affects activity on the surface of the sun.

Sunspot cycle is the pattern of solar activity where an average number of sunspot gradually increase and decrease.

Hence, the correct answer is option D

8 0
3 years ago
What is the pendulum length whose period is 2.0s ?
Mashutka [201]
Formula\ for\ period:\\\ T=2 \pi \sqrt{\frac{L}{g}}\\\ g-gravity=9,8 \frac{m}{s^2} ,\ L-pendulum \ length \\\\ \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\\\ \frac{T^2}{2 \pi } = \frac{L}{g} \\\\\ \frac{T^2}{2 \pi }*g=L\\\\ L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24mT=2 \pi   \sqrt{\frac{L}{g}} \\
 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
 \frac{T^2}{2 \pi }  = \frac{L}{g} \\
 \frac{T^2}{2 \pi }*g=L\\
L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m

7 0
3 years ago
If a system has 475 kcal of work done to it, and releases 5.00 × 102 kJ of heat into its surroundings, what is the change in int
Andrews [41]
First, we convert kcal to joules:
1 kcal = 4.184 kJ
475 kcal = 1987.4 kJ

Now, calculating the change in internal energy:

ΔU = Q + W; where Q is the heat supplied to the system and W is the work done on the system.

ΔU = -500 + 1987.4
ΔU = 1487.4 kJ
4 0
3 years ago
Read 2 more answers
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