Can someone please help me on this?

A car accelerates from a stand still to 75 m/s in 15 seconds what is the acceleration

inysia [295]

Answer: a = 75 m/s / second

Explanation:

8 0

2 years ago

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If a ball is thrown into the air with a velocity of 45 ft/s, its height in feet after t seconds is given by y=45t-15t2. Find the

Morgarella [4.7K]

Given that a ball is thrown into the air and its height in feet after t seconds is given by y = 45t - 15t², the instantaneous velocity when t=4 is -75ft/s.

<h3>What is the instantaneous velocity when t=4?</h3>

Given that;

Initial velocity = 45ft/s
Height in feet after t seconds is given by y = 45t - 15t²
Instantaneous velocity = ?

First, we determine the derivative of y = 45t - 15t²

y' = (1)45 - (2)15t

y' = 45 - 30t

We substitute in t = 4

y' = 45 - 30t

y' = 45 - 30(4)

y' = 45 - 120

y' = -75ft/s

Given that a ball is thrown into the air and its height in feet after t seconds is given by y = 45t - 15t², the instantaneous velocity when t=4 is -75ft/s.

Learn more about velocity here: brainly.com/question/2263547

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4 0

2 years ago

Two parallel wires carry equal currents in the opposite directions. Point A is midway between the wires, and B is an equal dista

zepelin [54]

Answer:

Complete Question:(missing part)

The current in the left wire having magnitude is = 210 A

The Current in the right wire having magnitude is = 10 A

Answer:

a. $B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )$

Ratio = B 1/B 2 = 210/10 = 21

$b.B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1}}{3}-I_{1}} )$

Ratio = B 1/B 2 = 210/3 X10 = 7

Explanation:

Ratio of the magnitude of the magnetic field at point A to that at point B =?

The current in the left wire having magnitude is = 210 A

The Current in the right wire having magnitude is = 10 A

distance between two wires = d

Given that both wires are carrying equal current but in opposite direction therefore

Magnetic field linked with the left wire at point A distance =d/2= according to biot savarts law

$B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{d}{2} }$

Magnetic field linked with the right wire at point B distance =d/2

$B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{d}{2} }$

Net magnetic field added

B net = B 1 + B 2

$B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (I_{1} +I_{2} )$

Ratio = B 1/B 2 = 210/10 = 21

Magnetic field linked to left wire at point B(d+ d/2)

$B = \frac{\mu_{0} \times I_{1} }{2\pi\times\frac{3d}{2} }$

Magnetic field linked to right wire at point B(d+ d/2)

$B = \frac{\mu_{0} \times I_{2} }{2\pi\times\frac{3d}{2} }$

Net magnetic field subtracted

B net = B 1 - B 2

$B net = \frac{\mu_{0} }{2\pi(\frac{d}{2}) } (\frac{I_{1} }{3}-1 } )$

Ratio = B 1/B 2 = 210/3 X10 = 7

4 0

3 years ago

A wave has a velocity of 24 m/s and a period of 3.0 s. Calculate the wavelength of the wave.

Katyanochek1 [597]

Velocity (unit:m/s) of the wave is given with the formula:

v=f∧,

where f is the frequency which tells us how many waves are passing a point per second (unit: Hz) and ∧ is the wavelength, which tells us the length of those waves in metres (unit:m)

f=1/T , where T is the period of the wave.

In our case: f=1/3

∧=v/f=24m/s/1/3=24*3=72m

5 0

3 years ago

An inexperienced catcher catches a 126 km/h fastball of mass 160 g within 1.36 ms, whereas an experienced catcher slightly retra

Firdavs [7]

Answer:

4117.65 N

Explanation:

Speed of ball, u = 126 km/h = 35 m/s

Mass of ball, m = 160 g = 0.16 kg

Time interval, t = 1.36 ms = 0.00136 s

We can calculate the force as a measure of the momentum of the ball:

F = P/t

Momentum, P, is given as:

P = mv

Therefore:

F = (mv) / t

F = (0.16 * 35) / (0.00136)

F = 4117.65 N

The force imparted to the two glove d hands of the inexperienced catcher is 4117.65 N.

6 0

3 years ago