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Harlamova29_29 [7]
2 years ago
15

Layla shot a balloon straight up in the air and it is in the air round trip for 6.80 s.

Physics
1 answer:
Katena32 [7]2 years ago
5 0

Answer:

24mph

Explanation:

it really depends how high but the average speed for that quick will be atleast 24mph if not try 42mph if it is wrong

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Why aren't iron, cobalt and iron in the same group of elements?
aev [14]
Because iron is a metal and cobalt is a non-metal
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3 years ago
35 POINTSS!!! PLSSSS HELLPPP!!!
Crazy boy [7]

Answer:

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beacuse:

Energy can be transferred from one object to another by doing work. ... When work is done, energy is transferred from the agent to the object, which results in a change in the object's motion (more specifically, a change in the object's kinetic energy).

6 0
3 years ago
A motorcycle is moving at 30 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.
shutvik [7]

Answer:

The motorcycle travelled 69.73 m during these 3.1 s.

Explanation:

In order to calculate the distance that the motorcycle travelled we first need to obtain the acceleration rate that was used to brake the vehicle. We do that by using the following formula:

a = (V_final - V_initial)/(t) = (15 - 30)/(3.1) = -4.84 m/s^2

The distance is given by the following formula:

S = (V_final^2 - V_initial^2)/(2*a)

S = (15^2 - 30^2)/[2*(-4.84)] = (225 - 900)/(-9.68) = -675/(-9.68) = 69.73 m

The motorcycle travelled 69.73 m during these 3.1 s.

5 0
3 years ago
Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0
3 years ago
A swimmer bounces almost straight up from a diving board and falls vertically feet first into a pool.she starts with a speed of
garik1379 [7]

Answer:

a) 1.20227 seconds

b) 0.98674 m

c) 7.3942875 m/s

Explanation:

t = Time taken

u = Initial velocity = 4.4 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow 0=4.4-9.81\times t\\\Rightarrow \frac{-4.4}{-9.81}=t\\\Rightarrow t=0.44852\ s

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.4\times 0.44852+\frac{1}{2}\times -9.81\times 0.44852^2\\\Rightarrow s=0.98674\ m

b) Her highest height above the board is 0.98674 m

Total height she would fall is 0.98674+1.8 = 2.78674 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.78674=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.78674\times 2}{9.81}}\\\Rightarrow t=0.75375\ s

a) Her feet are in the air for 0.75375+0.44852 = 1.20227 seconds

v=u+at\\\Rightarrow v=0+9.81\times 0.75375\\\Rightarrow v=7.3942875\ m/s

c) Her velocity when her feet hit the water is 7.3942875 m/s

3 0
3 years ago
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