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Allushta [10]
1 year ago
10

a rocket with a mass of 4kg accelerates up from the ground at a rate of 20ft m/s^2, what is the drag force acting on the rocket

if the thrust of the rockets is 150n
Physics
1 answer:
Maru [420]1 year ago
8 0

The drag force acting on the rocket is 80N.

<h3>Give an explanation of drag force?</h3>

The divergence in velocity between the fluid and the item, also known as drag, exerts a force on it. Between the liquid and the solid object, there should be motion. Drag is absent in the absence of motion.

The air molecules are more compressed (pushed together) on the surfaces that are facing the front while being more dispersed (spread out) on the surfaces facing the back. Turbulent flow, which occurs when air layers split from the surface and start to swirl, is what causes this.

The drag force acting on the rocket F = ma

Given,

m = 4kg, a = 20ftm/s²

Substituting m and a values in the above formula,

The drag force acting on the rocket F = 4×20

The drag force acting on the rocket F = 80N.

To know more about drag force visit:

brainly.com/question/15144984

#SPJ4

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A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum
Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , m_1= 0.473 kg

mass of puck 2 , m_2= 0.819 kg

initial speed of puck 1 , u_1=2.76\frac{m}{s}

initial speed of puck 2 , u_2=0.00\frac{m}{s}

Final speed of puck 1 and puck 2 be v_1\, and\, v_2  respectively

Apply conservation of linear momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

=>0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2

=>1.594=0.5775\times v_1+ v_2 -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

u_2-u_1=v_1-v_2

=>0-2.76=v_1-v_2 ------(B)

From equation (A) and (B)

v_1=-0.739\frac{m}{s}

and v_2=2.02\frac{m}{s}

Thus the final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

       

3 0
3 years ago
an airplane releases a ball as it flies parallel to the ground at a height of 235m. if the ball lands on the ground exactly at 2
Oksanka [162]
<span>When the question says the ball lands a distance of 235 meters from the release point, we can assume this means the horizontal distance is 235 meters. Let's calculate the time for the ball to fall 235 meters to the ground. y = (1/2)gt^2 t^2 = 2y / g t = sqrt{ 2y / g } t = sqrt{ (2) (235 m) / (9.81 m/s^2) } t = 6.9217 s We can use the time t to find the horizontal speed. v = d / t v = 235 m / 6.9217 s v = 33.95 m/s Since the horizontal speed is the speed of the plane, the speed of the plane is 33.95 m/s</span>
7 0
3 years ago
F = 50 N<br> m = 72 kg<br> m/s2
lyudmila [28]

Answer:

Explanation:

F=ma

a=F/m

a=50/72=

a=0.694

3 0
3 years ago
In a $100$ meter track event, Alice runs at a constant speed and crosses the finish line $5$ seconds before Beatrice does. If it
Liono4ka [1.6K]

Answer:

10s

Explanation:

If it took Beatrice 25 seconds to complete the race

Distance = 100 meter

Beatrice speed = 100/25

                          = 4m/s

If Alice runs at a constant speed and crosses the finish line $5$ seconds, she must have completed the race in 20s (25 -5).

Her speed where constant

= 100/20

= 5 m/s

It would take Alice

= 50/5

= 10s

It would take Alice 10s to run $50$ meters.

5 0
3 years ago
Find the minor measurement of the vernier scale by taking 49, 1mm divisions of the main scale and dividing it into 50 vernier di
olga nikolaevna [1]
<h2>♨ANSWER♥</h2>

length of V-50 = 49mm

length of V-1 = 49/50mm

= 0.98mm

so,

minor measurement = (M-1) - (V-1)

= 1mm -0.98mm

= 0.02mm

☆ Therefore,

The minor measurement of the vernier scale is 0.02mm.

<u>☆</u><u>.</u><u>.</u><u>.</u><u>hope this helps</u><u>.</u><u>.</u><u>.</u><u>☆</u>

_♡_<em>mashi</em>_♡_

6 0
2 years ago
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