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2 years ago

5

A cricket can travel approximately 8 m/s. How many meters could a cricket travel in 75 s?

1 answer:

nika2105 [10]2 years ago

7 0

Answer:

Your answer will be 600meters

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FinnZ [79.3K]

Answer:

A hope this helps

Explanation:

8 0

2 years ago

A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below t

maria [59]

Answer:

Explanation:

Suppose v is the initial velocity and $\theta$ is the angle of inclination

distance traveled in vertical direction in t=1 s

When gravity is present

$y=vt+\frac{1}{2}at^2$

where $y=vertical\ distance$

$a=acceleration$

$t=time$

$v=initial\ velocity$

here initial velocity is v\sin \theta [/tex] so

$y=v\sin \theta \times 1-\frac{1}{2}gt^2$

$y=v\sin \theta -0.5g$

In absence of gravity

$y_2=v\sin \theta \times t$

$y_2=v\sin \theta \times 1$

$\Delta y=y_2-y=v\sin \theta -v\sin \theta +0.5 g=4.9\ m$

4 0

2 years ago

In which way are the cathode rays deflected in the x-plates of the Cathode Ray Oscilloscope

MrRissso [65]

Answer:

The cathode ray is deflected vertically to the fluorescent screen

Explanation:

$.$

4 0

2 years ago

Read 2 more answers

How far would a jet going 155 m/s travel in 9 s?

docker41 [41]

Answer:

1,395 m

Explanation:

155×9

multiply m/s by 9s

5 0

3 years ago

The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it

Keith_Richards [23]

Answer:

The no. of revolutions does the tub turn while it is in motion is = 52.51 revolutions

Explanation:

Given data

$\omega_1$ = 0

$\omega_2$ = 5 $\frac{rev}{sec}$

Time taken = 7 sec

(1). The angular acceleration is given by

$\alpha = \frac{d \omega}{dt}$

$\alpha = \frac{5}{7}$

$\alpha = 0.714 \frac{rev}{s^{2} }$

We know that from the equation of motion

$\omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_1$

$5^{2} = 0 + 2 (0.714) \theta_1$

$\theta_1 = 17.5 \ rev$ -------- (1)

(2). The angular acceleration is given by

$\alpha = \frac{d \omega}{dt}$

$\alpha = - \frac{5}{14}$

$\alpha =$ - 0.357 $\frac{rev}{s^{2} }$

We know that from the equation of motion

$\omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_2$

$0^{2} = 5^{2} + 2 (-0.357) \theta_2$

$\theta_2$ = 35.01 rev ------- (2)

Total no of revolution made by the machine is

$\theta = \theta_1 + \theta_2$

$\theta =$ 17.5 + 35.01

$\theta =$ 52.51 rev

Therefore the no. of revolutions does the tub turn while it is in motion is = 52.51 rev

8 0

2 years ago