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2 years ago

If a 15 kg mass accelerates at a rate of 4 m/s2, what net force acts on it?

Physics

1 answer:

jarptica [38.1K]2 years ago

5 0

<h2>Answer: D 60N</h2>

<h3>Explanation:</h3>

Mass(M)=15 kg

Acceleration(A)=4 m/s2

Force=?

Now,

Force(F)=M×A

F=15×4

F=60N Ans

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The function of a cell depends primarily on its ?

Schach [20]

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3 years ago

7. Katie and her best friend liam play tennis every saturday morning. When katie serves the ball to liam, it travels 9.5 meters

Ksenya-84 [330]

Answer:

The velocity of the tennis ball is 4.52 m/s.

Explanation:

Given that,

The distance covered by ball, d = 9.5 meters (due south)

Time, t = 2.1 sec

Let v is the velocity of the tennis ball. We know that the velocity of an object is given by the total distance covered divided by total time taken. It is given by :

$v=\dfrac{d}{t}\\\\v=\dfrac{9.5\ m}{2.1\ s}\\\\v=4.52\ m/s$

So, the velocity of the tennis ball is 4.52 m/s. Hence, this is the required solution.

6 0

3 years ago

A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th

Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

$E=-\frac{d}{dt}(BACOS\alpha )\\$

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using $\pi r^{2} \\$ where r is the radius of 5cm or 0.05m

$A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\$

since we no that the angle is at $0^{0}$

we determine the magnitude of the magnetic filed

$B=0.5e^{-t} \\t=2s$

$E=-(0.5e^{-2} * 0.00785)$

$E=-0.000532v\\$

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

$V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A$

$I=2.6mA$

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3 years ago

The compressor on an air conditioner draws 66.1 a when it starts up. the start-up time is about 0.388 s. how much charge passes

Ierofanga [76]

The current intensity is defined as the amount of charge flowing through a certain point of a wire divided by the time interval:
$I= \frac{Q}{\Delta t}$
where Q is the charge and $\Delta t$ is the time. Re-arranging the formula, we have
$Q=I \Delta t$

for the compressor in our problem, the intensity of current is I=66.1 A, while the time is $\Delta t= 0.388 s$ , so the amount of charge that crosses a certain point of the circuit during this time is
$Q=I \Delta t=(66.1 A)(0.388 s)=25.65 C$

5 0

3 years ago

a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second

lawyer [7]

Answer:

a) Yes, there are $10^8$ shakes in a second ( $1 s \frac{1 shake}{10^{-8}s}=10^8 shake$ ) in a year there are 31,536,000 seconds... that is 3.1536x $10^7$ ( $1year\frac{365 day}{1 year} \frac{24 h}{1 day} \frac{60 min}{1 h}\frac{60s}{1min}=3.1536 *10^7$ )

b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

$106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s$

That is about 2 and half hours.

3 0

3 years ago