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Ivahew [28]
3 years ago
12

9. Which is the type of force that causes acceleration?​a.balancedb.zeroc.unbalanced

Physics
1 answer:
Artist 52 [7]3 years ago
6 0

Explanation:

<h3>Answer:-</h3>

<u>Unbalanced</u><u> </u>Force causes acceleration.

But why?

This is because for accelerating a body we need unbalanced forces to move it.

Hope it helps :)

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When driving downhill, your vehicle will accelerate and start moving fast, due to _______________.
Stolb23 [73]
B: Gravity.
The force of gravity will pull the car down the hill. The weight/mass of the car also helps this.
3 0
4 years ago
Read 2 more answers
A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pu
vredina [299]

Answer:

a) v_1=5.5172\ m.s^{-1}

b) v_2=-1.5152\ m.s^{-1}

c) v_3=4.6154\ m.s^{-1}

d) v_{avg}=3.8462\ m.s^{-1}

Explanation:

Given:

  • distance down the field in the first interval, d_1=16\ m
  • time duration of the first interval, t_1=2.9\ s
  • distance down the field in the second interval, d_2=-2.5\ m
  • time duration of the second interval, t_2=1.65\ s
  • distance down the field in the third interval, d_3=24\ m
  • time duration of the third interval, t_3=5.2\ s

a)

velocity in the first interval:

v_1=\frac{d_1}{t_1}

v_1=\frac{16}{2.9}

v_1=5.5172\ m.s^{-1}

b)

velocity in the second interval:

v_2=\frac{d_2}{t_2}

v_2=\frac{-2.5}{1.65}

v_2=-1.5152\ m.s^{-1}

c)

velocity in the third interval:

v_3=\frac{d_3}{t_3}

v_3=\frac{24}{5.2}

v_3=4.6154\ m.s^{-1}

d)

We know that the average velocity is given as the total displacement per unit time.

v_{avg}=\frac{d_1+d_2+d_3}{t_1+t_2+t_3}

v_{avg}=\frac{16-2.5+24}{2.9+1.65+5.2}

v_{avg}=3.8462\ m.s^{-1}

3 0
3 years ago
Block M = 7.50 kg is initially moving up the incline and is increasing speed with a = [09]____________________ m/s2 . The applie
Aliun [14]

Answer:

a) 73.2N

b) 66.6N

c) 20.28N

Explanation:

F= mg=7.5×9.8=73.5N

Force parallel to the plane Fp= 73.5sin25= 31.06N

b) Fv= 73.5 cos25= 66.6N

c) Ff= u×Fv= 0.312×66.6=20.28N

a) Normal ForceFn= F/(cos25) - Fp - Ff = ma

1.1F -31.06-20.28=7.5×3.82

1.1F -51.84=28.65

1.1F= 28.65+51.84

F= 80.49/1.1

F= 73.2N

4 0
3 years ago
What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m
masya89 [10]

Answer: 0.0058 m/s^{2}

Explanation:

Centripetal acceleration a_{C} is calculated by the following equation:

a_{C}=\frac{V^{2}}{r}

Where:

V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s is the Earth's orbital speed

r=1.5(10)^{11} m is the orbital radius

a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}

a_{C}=0.0058 m/s^{2}

4 0
3 years ago
The Pacific Plate is moving 29 mm/year toward the north and 20 mm/year toward the west relative to the North American Plate. Sho
adell [148]

Answer:

7 mm per year

Explanation:

It is given that :

The Pacific plate is moving towards north at = 29 mm per year

The Pacific plate is moving towards west at = 20 mm per year

We have to calculate the total relative motion towards the northwest.

So we have to find the resultant of the two motions.

Since the two movements are perpendicular, therefore the angle between the two motions is 90 degree.

Therefore, finding their resultant,

R^2=(29)^2+(20)^2

R^2=(49)^2

R = 7

Therefore, total relative motion towards the northwest is 7 mm per year.

   

4 0
3 years ago
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