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krok68 [10]
2 years ago
9

This graph indicates that the acceleration between 2.00 and 4.00 seconds is the same as that between 12.00 and 14.00 seconds.

Physics
1 answer:
adoni [48]2 years ago
5 0
I believe it’s false I’m probably wrong I ju at need points for my test
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6. During an impact time casting 5 x 10-45 a gulf club exerts an average impar
Novay_Z [31]

Answer:

2.5 × 10-⁴¹ Ns

Explanation:

Impulse

I = F × t

I = 5000 N × 5 × 10-⁴⁵ s

I = 25 × 10-⁴² Ns

I = 2.5 × 10-⁴¹ Ns

#LearnWithEXO

6 0
3 years ago
Solve for the length of the inclined plane if the angle equals 19.45 degrees.
mel-nik [20]

The length of the inclined plane is approximately 12 ft

The situation forms a right angle triangle.

<h3>Right triangle</h3>

Right triangle have one of its angle as 90 degrees.

Therefore,

The length of the inclined plane is the hypotenuse of the triangle. The length of the inclined plane can be found using trigonometric ratios.

height = 4 ft

angle(∅) = 19.45°

sin 19.45 = 4 / h

h = 4 / 0.33298412235

h = 12.0125847796

h = 12 ft

Therefore, the length of the inclined plane is approximately 12 ft

learn more on inclined plane:brainly.com/question/14163589?referrer=searchResults

5 0
2 years ago
Does anyone know this?
e-lub [12.9K]
Explanation: The first one
Source: it literally has fusion in the name
5 0
2 years ago
Read 2 more answers
A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of th
Andrew [12]

Answer:

Explanation:

A )

At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy

1/2 m V² = mg x 2r + 1/2 mv²

m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top

1/2  V² = g x 2r + 1/2 v²

 V² = g x 4r +  v²

 V² = 9.8 x 4 +  8²

V² = 103.2

V = 10.16 m/s

B )

If T be the tension at the top

Net downward force

= mg + T . This force provides centripetal force for the circular motion

mg +T = mv² / r

T =   mv²/r -mg

= m ( v²/r - g )

= .005 ( 8²/1 -g )

= .005 x 54.2

= .27 N .

C ) At the bottom

Net force = T  - mg , T is tension at the bottom , V is velocity at bottom

T-mg = mV²/r

T = m ( V²/r +g )

= .005 ( 10.16²/1 +9.8)

= .005 x 113

= .56 N .

3 0
3 years ago
Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject
Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}

where

\mu_{sun} is gravitational parameter of sun =  1.32712 x 10^20 m^3 s^2.t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}

t = 12,105.96 sec

6 0
3 years ago
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