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damaskus [11]
3 years ago
14

Jen falls out of a tree and lands on a trampoline. The trampoline sags 60 cm before launching Jen back into the air. At the very

bottom, where the sag is the greatest, is Jen’s acceleration upward, downward, or zero?
Physics
1 answer:
Ray Of Light [21]3 years ago
7 0

Answer:

At the very bottom, whnere the sag is the greatest, Jay’s acceleration is upward.

Explanation:

As Jay lands on the trampoline, Jay’s motion was dowward,  the trampoline was opposing his motion and hence, caused him to reach an initial halt position. Afterwards, the trampoline causes Jay to move back into the  air and therefore, the change in velocity vector act in upward direction. The acceleration vector is always align towards the change in velocity vector's direction.

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After their loss to U of A, the ASU sumdevils traveled back to Tempe which is 187
Dmitry_Shevchenko [17]

The average speed is 20.8 m/s

Explanation:

The average speed for the trip is given by:

v=\frac{d}{t}

where

d is the distance covered

t is the time elapsed

For the trip in this problem, we have:

d = 187 km = 187,000 m is the distance travelled

The initial  time is 10:00 pm while the arriving time is 12:30 am: this means that the time elapsed is 2.5 hours. Converting into seconds,

t = 2.5 h \cdot (60)(60)=9000 s

Therefore, the average speed for the trip is

v=\frac{187,000}{9000}=20.8 m/s

Learn more about average speed:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

5 0
3 years ago
What is the difference in
andreev551 [17]

Answer:

0.438kg/ms-¹

Explanation:

Momentum, denoted by p, can be calculated by using the formula;

p = mv

Where;

m = mass (kg)

v = velocity (m/s)

Momentum (p) of bird = 0.216 kg × 5.87 m/s = 1.268kg/ms-¹

Momentum (p) of crawling baby = 7.29 kg kg × 0.234 m/s = 1.706kg/ms-¹

Having calculated the momentum of the bird to be 1.268kg/ms-¹, and the momentum of the baby to be 1.706kg/ms-¹, the difference in momentum between the flying bird and the crawling baby is:

{1.706kg/ms-¹ - 1.268kg/ms-¹} = 0.438kg/ms-¹

6 0
2 years ago
Read 2 more answers
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
2 years ago
To control whether an object is solid or incorporeal (things can pass through it) you would use the:
Zarrin [17]

Answer:

Gamma radiation or Cathode rays

Explanation:

by striking incident gamma or cathode rays onto the solid when placed on a photographic plate

5 0
2 years ago
Three cannons are located at the top of a cliff above a level plain. Cannon A is aimed at an angle of 25° above the horizontal a
vodka [1.7K]

Answer:

The canon B hits the ground fast.

Explanation:

Given that,

Speed of cannon A = 85 m/s

Speed of cannon B= 100 m/s

Speed of cannon C = 75 m/s

We need to calculate the cannonballs will hit the ground with the greatest speed

Using conservation of energy

The final kinetic energy of canon depends on initial kinetic energy and potential energy.

The  final velocity depends upon initial velocity and initial height.

So,  the initial velocity of canon B is high.

Hence, The canon B hits the ground fast.

3 0
3 years ago
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