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4 years ago

A car travels from rest to 55 m/s in 10 seconds. What is its acceleration?

Physics

1 answer:

maria [59]4 years ago

5 0

Answer:

(55-0)/10 --> 55 m/s / 10 s --> 5.5 m/s^2

Explanation:

I WANT U TO PARK THAT BIG MACK TRUCK, RIGHT IN THIS LITTLE GRAGE

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Nellie pulls on a 10kg wagon with a constant horizontal force of 30N. If there are no other horizontal forces, what is the wagon

s2008m [1.1K]

Answer:

The acceleration of the wagon is 3 m/s².

To calculate the acceleration of the wagon, we use the formula below.

Formula:

F = ma............. Equation 1

Where:

F = horizontal Force

m = mass of the wagon

a = acceleration of the wagon.

make a the subject of the equation

a = F/m.............. Equation 2

From the question,

Given:

F = 30 N

m = 10 kg

Substitute these values into equation 2

a = 30/10

a = 3 m/s²

Hence, the acceleration of the wagon is 3 m/s².

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2 years ago

What's a good way to determine the net force of something

ollegr [7]

By adding up all the individual forces of the object

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3 years ago

If an R = 1-kΩ resistor, a C = 1-μF capacitor, and an L = 0.2-H inductor are connected in series with a V = 150 sin (377t) volts

fgiga [73]

Answer

given,

R = 1-kΩ = 1000 Ω

C = 1-μF

L = 0.2-H

V = V_max sin( ω t)

comparing

V = 150 sin ( 377 t)

ω = 377

$\chi_c = \dfrac{1}{\omega C}$

$\chi_c = \dfrac{1}{377 \times 1 \times 10^{-6}}$

$\chi_c = 2652.5\Omega$

$\chi_L =377 \times 0.2$

$\chi_L =75.4\ \Omega$

Impedance,

$Z = \sqrt{R^2+(\chi_L-\chi_c)^2}$

$Z = \sqrt{1000^2+(75.4 -2652.5)^2}$

Z = 2764.3 Ω

now,

V_{max} = 150 V

$I_{max} = \dfrac{V}{Z}$

$I_{max} = \dfrac{150}{2764.3}$

$I_{max} = 0.0543$

$I_{max} = 54.3\ mA$

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3 years ago

Advantage of using infared camera

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3 years ago

Read 2 more answers

A centrifuge accelerates uniformly from rest to 15000 rpm in 330 s . Through how many revolutions did it turn in this time?

eduard

Answer:

The number of revolutions turned by the centrifuge is 8250 revolutions.

Explanation:

Given;

number of revolution per minutes, ω = 15000 rpm

time of motion, t = 330 s = 5.5 minutes

The number of revolutions turned by the centrifuge is given by;

$N = \frac{1500 \ Rev}{minutes} *5.5 \ minutes\\\\N = 8250 \ revolutions$

Therefore, the number of revolutions turned by the centrifuge is 8250 revolutions.

4 0

3 years ago