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3 years ago

Suppose a balloon was tied to a pole. What happens?

Physics

1 answer:

Elodia [21]3 years ago

8 0

Answer:

it wont fly away

Explanation:

depending how tight the knott and find it is going to be stuck to the pole

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Conduction can only happen when objects are in:

serious [3.7K]

Answer:

When it’s gravitational

Explanation:

I’m pretty sure that’s the answer

5 0

3 years ago

Read 2 more answers

A can of beans that has mass M is launched by a spring-powered device from level ground. The can is launched at an angle of α0 a

scZoUnD [109]

Hi there!

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
$T_h = \frac{T}{2}$

Now, we can determine the velocity at which the can was launched at using the following equation:
$v_f = v_i + at$

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.

Therefore:
$0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}$

***vsinθ is the vertical component of the velocity.

Solve for 'v':
$vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}$

Now, recall that:
$W = \Delta KE = \frac{1}{2}m(\Delta v)^2$

Plug in the expression for velocity:
$W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}$

We can use the same process as above, where T' = 2T and Th = T.

$v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}$

The work done in part B is 4 times greater than the work done in part A.

$\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}$

4 0

2 years ago

3) All numbers that are divisible by both 3 and 5 are also divisible by 10. Which of the following numbers can be used to show t

zlopas [31]

Answer:

answer E

Explanation:

cuz its 45 is divisible by 3 and 5 but not 10

8 0

3 years ago

Two vehicles approach an intersection: a truck moving eastbound at 16.0 m/s and an SUV moving southbound at 20.0 m/s. Suppose th

mario62 [17]

Answer:25.61 m/s

Explanation:

Given

truck is moving eastbound with a velocity of 16 m/s

Velocity of truck $v_t=16\hat{i}$

SUV is moving south with a velocity of 20 m/s

Velocity of SUV in vector form $v_s=-20\hat{j}$

Velocity of truck relative to the SUV

$v_{ts}=v_{t}-v_s$

$v_{ts}=16\hat{i}-(-20\hat{j})$

Magnitude of relative velocity is

$|v_{ts}|=\sqrt{16^2+20^2}$

$|v_{ts}|=25.61\ m/s$

5 0

3 years ago

A car moving at a steady 10 m/s on a level highway encounters a bump that has a circular cross-section with a radius of 30 m. Th

daser333 [38]

When you are driving over a circular bump, you can feel your body moving up a small distance in the seat. This make you feel like you weigh less that you really do. This is caused by the fact that net down force on you is equal to the centripetal force. If you go the following website, you will see a problem that is similar to this one. Go the problem called Sample Roller Coaster Problem.

At this website you can see that the normal force exerted by the seat of the car on a 60.0-kg passenger is equal to the difference of the passenger’s weight and the centripetal force.

Weight = 60 * 9.8 = 588 N

Fc = 60 * 10^2/30 = 200 N

Normal force = 388 N

Explanation:

8 0

3 years ago