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NARA [144]
3 years ago
7

Which quantities decrease as the distance between a planet and the Sun increases? Check all that apply.

Physics
2 answers:
sweet-ann [11.9K]3 years ago
5 0

<u>Answer</u>

Gravitational force

<u>Explanation</u>

Gravitational force is the force of attraction between two bodies of a given masses. It is calculated as follows:

Gravitational force = GM₁M₂/r²

From the formula it can be seen that gravitational force is inversely proportional to the r where are is the distance between the two bodies.

When the distance increases the gravitational force decreases.

Zielflug [23.3K]3 years ago
4 0

Here’s a picture of the correct answers.

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Anaya Goodloe decides to spike the football after a big touchdown (on Earth) and it
Llana [10]

Answer:

The football leaves with the velocity, u = 15.68 m/s

Explanation:

Given data,

The football bounces back up off the ground and is airborne for, t = 3.2 s

Let the football bounces back up off the ground in the vertical direction

The formula for time of flight is given by,

                              t = 2u /g

∴                             u = gt / 2

Substituting the values,

                               u = 9.8 x 3.2 / 2

                                u = 15.68 m/s

Hence, the football leaves with the velocity, u = 15.68 m/s

6 0
3 years ago
Does Mike wosowski blink or wink
aev [14]

Answer:

both

Explanation:

7 0
3 years ago
Read 2 more answers
An airplane flies in a loop (a circular path in a vertical plane) of radius 160 m . The pilot's head always points toward the ce
notka56 [123]

Answer:

a) 39.6 m/s b) 4123 N

Explanation:

a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).

Fnet=ma

ma=m(v^2/R) (centripetal acceleration)

mg=m(v^2/R)

m cancels out (this is why pilot feels weightless) so,

g=(v^2/R)

9.8 m/s^2 = v^2/160 m

v^2=1568 m^2/s^2

v=39.6 m/s

b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.

Convert 300 km/hr to m/s

300 km/hr=83.3 m/s

Convert pilot's weight into mass:

760 N = 77.55 kg

Fnet=ma

n-mg=m(v^2/R)

n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)

n=3363.2 N+760 N=4123 N

5 0
3 years ago
If a 10kg block is at rest on a table and a 1200N force is applied in the eastward direction for 10 seconds, what is the acceler
gavmur [86]

Answer:

a = 120 m/s²

Explanation:

We apply Newton's second law in the x direction:

∑Fₓ = m*a Formula (1)

Known data

Where:

∑Fₓ: Algebraic sum of forces in the x direction

F: Force in Newtons (N)

m: mass (kg)

a: acceleration of the block (m/s²)

F = 1200N

m = 10 kg

Problem development

We replace the known data in formula (1)

1200 = 10*a

a = 1200/10

a = 120 m/s²

6 0
3 years ago
A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th
melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0
3 years ago
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