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3 years ago

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the gravitational pull of the moon is much less than the gravitational pull of earth, which two statements are true for an objec

t with a mass of 20 kilograms that weighs 44 pounds on earth

1 answer:

alexgriva [62]3 years ago

6 0

Answer:

b

Explanation:

the earths mass is more than the moon .

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A student, standing on a scale in an elevator at rest, sees that his weight is 840 n. as the elevator rises, his weight increase

Firlakuza [10]

<span>1) at rest his weight is 840 N

=> 840N = mass * g => mass = 840 N / g = 840 N / 9.8 m/s^2 = 85.7 kg

2) as the elevator rises, his weight increases to 1050 N,

The reading of the scale is the norma force of it over the body of the person.

And the equation for the force is: Net force = mass * acceleration = normal force - weight at rest

=> mass * acceleration = 1050 N - 840 N = 210 N

acceleration = 210 N / mass = 210 N / 85.7 kg = 2.45 m/s^2 (upward)

3) when the elevator slows to a stop at the 10th floor, his weight drops to 588 N

=> mass * acceleration = 588 N - 840 N = - 252 N

=> acceleration = - 252 N / 85.71 kg = - 2.94 m / s^2 (downward)

Answer:

Acceleration at the beginning of the trip 2.45 m/s^2 upward
Acceleration at the end of the trip 2.94 m/s^2 downward
</span>

6 0

3 years ago

A 60.0-kg skater begins a spin with an angular speed of 6.0 rad/s. By changing the position of her arms, the skater decreases he

juin [17]

The skater's final angular speed is equal to 12 rad/s.

When implemented to angular momentum, the regulation of conservation means that the momentum of a rotating item is no longer exchanged until some form of external torque is carried out. Torque, in this sense, can check with any outside pressure that acts upon the object for the purpose to twist or rotate.

The law of conservation of angular momentum states that once no external torque acts on an item, no trade of angular momentum will occur. The angular momentum of a machine is conserved as long as there may be no net external torque performing on the machine.

In angular kinematics, the conservation of angular momentum refers back to the tendency of a device to keep its rotational momentum inside the absence of outside torque. For a round orbit, the system for angular momentum is (mass) ×(pace) ×(radius of the circle): (angular momentum) = m × v × r.

Learn more about angular momentum here brainly.com/question/7538238

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7 0

1 year ago

Assume a beam of light hits the boundary separating medium 1, with index of refraction n1 and medium 2, with index of refraction

Reptile [31]

Answer:

option C

Explanation:

Given,

Refractive index of medium 1 = n₁

Refractive index of medium 2 = n₂

For total internal reflection to take place light should move from denser medium to the rarer medium.

Here Total internal reflection take place at the boundary of medium 1 and medium 2 so, the refractive index of medium 1 is more than medium 2

n₁ > n₂

The correct answer is option C

3 0

3 years ago

How fast would you be going (in kmh) if you had a ship that accelerated at a constant 1g for 24 hours?

Nady [450]

Answer:

Explanation:

1 g is 9.8 m/s^2 the problem wants the results in km/h so we'll fix that really quick.

9.8 m/s^2 (1 km/1000m)(60 sec/1 min)^2(60 min/1 hour)^2 = 127008 km/hour^2

Now, I'm assuming the ship is starting from rest, and hopefully you know your physics equations. We are going to use vf = vi + at. Everything is just given, or we can assume, so I'll just solve.

vf = vi + at

vf = 0 + 127008 km/hour^2 * 24 hours

vf = 3,048,192 km/hour

If there's anything that doesn't make sense let me know.

5 0

3 years ago

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

6 0

2 years ago