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Novosadov [1.4K]

3 years ago

14

A hot-air balloon is rising straight up with a speed of 1.30 m/s. A ballast bag is released from rest relative to the balloon wh

en it is 5.36 m above the ground. How much time elapses before the ballast bag hits the ground?

1 answer:

DiKsa [7]3 years ago

6 0

Answer:

The time before the ballast hits the ground is 1.186918s

Explanation:

We assume that the upward direction is positive.

The bag released from the ballon is at rest. The velocity of the bag equals the velocity of the balloon.

The balloon has a velocity (V0y) = 3m/s

When the bag is released, it has a height of 5.36m

To touch the ground it makes a displacement of y = - 5.36m

The motion of the bag can be described as followed:

y = V0t + 1/2 * at²

t² + (2V0/a)*t - (2y/a) = 0

We can solve this equation for t:

t= ((-2V0/a) ± √((2V0/a)² + 4*(2y/a))) /2

t = -(V0/a) ± √((V0/a)² + (2y/a))

In this equation we can plug the given values ( V0 = 1.3 m/s ; y = 5.36m ; g =-9.8 m/s²)

t = (1.3/-9.8)±√((1.3/-9.8)²+(2*(-5.36)/-9.8))

t = 0.132653 ± 1.054266

t= 1.186918

The time before the ballast hits the ground is 1.186918s

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valentinak56 [21]

Answer:

$F=6\times 10^{-7}\ N$

Explanation:

Given:

quantity of point charge, $q=-4\times 10^{-9}\ C$

radial distance from the linear charge, $r=0.09\ m$

linear charge density, $\lambda=3\times 10^{-9}\ C.m^{-1}$

<u>We know that the electric field by the linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.\epsilon_0.r}$

$E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}$

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$F=E.q$

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$F=6\times 10^{-7}\ N$

3 0

4 years ago

The length of the side of a cube is 10.0 ±0.3 cm. What is the absolute uncertainty in the perimeter of one side of the cube?

Nimfa-mama [501]

Answer:Stuff

Explanation: Nun

4 0

3 years ago

In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn

kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

$s=ut+\dfrac{1}{2}at_{A}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times1.6\times t_{A}^2$

$t_{A}=\sqrt{\dfrac{30\times2}{1.6}}$

$t_{A}=6.12\ sec$

We need to calculate the time for B

Using equation of motion

$s=ut+\dfrac{1}{2}at_{B}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times2.0\times t_{B}^2$

$t_{B}=\sqrt{\dfrac{30\times2}{2.0}}$

$t_{B}=5.48\ sec$

We need to calculate the time for longest nap

Using formula for difference of time

$t'=t_{A}-t_{B}$

$t'=6.12-5.48$

$t'=0.64\ s$

Hence, B can take 0.64 sec for the longest nap .

4 0

4 years ago

andrew-mc [135]

Answer:

V=W/Q

107V= W/17C

= We= 107×17 J

= 1819 J

Explanation:

hope it helps

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3 years ago

The photograph shows matter going through a physical change. Which characteristic of the matter changes?

Lapatulllka [165]

Answer:

Possibly shape.

Explanation:

Because it is ripped in half. As you guys and the questioner can see that there is a wiggly strip down the middle. That also cause of a shape change.

I really hope my answer is right ^_^

7 0

3 years ago

Read 2 more answers