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antiseptic1488 [7]

3 years ago

10

3. A car going 22m/s accelerates to pass a truck. Five seconds late the car is going 35m/s. Calculate the acceleration of the ca

r.

1 answer:

DedPeter [7]3 years ago

6 0

Answer:

57

Explanation:

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What is the minimum speed must a salmon jumping at an angle of 35.2 leave the water in m/s?

yuradex [85]

Height of the waterfall is 0.449 m

its horizontal distance will be 2.1 m

now let say his speed is v with which he jumped out so here the two components of his velocity will be

$v_x = vcos35.2 = 0.817 v$

$v_y = vsin35.2 = 0.576 v$

here the acceleration due to gravity is 9.81 m/s^2 downwards

now we can find the time to reach the other end by y direction displacement equation

$\Delta y = v_y * t + \frac{1}{2} at^2$

$-0.449 = 0.576 * v *t - \frac{1}{2}*9.81 * t^2$

also from x direction we can say

$\Delta x = v_x * t$

$2.1 = 0.817 v* t$

now we have

$v* t = 2.57$

we will plug in this value into first equation

$- 0.449 = 0.576 * 2.57 - 4.905 * t^2$

$1.93 = 4.905 * t^2$

$t = 0.63 s$

now as we know that

$v* t = 2.57$

t = 0.63 s

$v = \frac{2.57}{0.63}$

$v = 4.1 m/s$

so his minimum speed of jump is 4.1 m/s

8 0

3 years ago

What is the speed of a 400 kg elevator that has 1800J of energy

sveta [45]

Answer: 1400

Explanation:

8 0

2 years ago

What are the three primary particles that make up an atom and what are associated with electromagnetic charges

Sonbull [250]

Answer:

It's (c). ..........................

8 0

3 years ago

A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 10.0 m

vovangra [49]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

Angular velocity is 1.3823 rad/s

Linear velocity is given by

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

Centripetal force is given by

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

7 0

2 years ago

A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.7 seconds, coasts for 2.8 s , and then slows down at a r

Lisa [10]

Answer:

18.43m

Explanation:

At the initial stage, the car accelerates at 2.0 m/s² for 6.7 seconds, to get the velocity of the car at this point, we will use the equation of motion;

v = u + at

v = 0+(2.0)(6.7)

v = 13.4m/s

The velocity of the car during this time is 13.4m/s.

Also, if the car slows down at a rate of 1.5m/s², we can also calculate the time it took for the car to decelerate (check the attachment for diagram).

v = u+at

v = 0

u = 13.4m/s

a = -1.5m/s² (deceleration is negative acceleration)

0 = 13.4+-1,5t

-13.4 = -1.5t

t = 13.4/1.5

t = 8.93s

Hence it took the car 8.93s to slow down for the next stop sign.

To calculate how far apart the stop signs are, we need to calculate the total distance AD according to the diagram

Distance covered = AD = 18.43m

HENCE THE STOP SIGNS ARE 18.43m apart

4 0

3 years ago