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3 years ago

An object, initially at rest moves 250m in 17s. What is it's acceleration?

1 answer:

8 0

With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity

d= 250s
t= 17s

a= d/t
= $\frac{250}{17}$
= 4.7 $\frac{m}{s}$

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Find the sum 5.24 g, 43.261 g, and 7.3458 g. Write your answer with the correct amount of significant figures.

lapo4ka [179]

Answer:

This is how I do it:

5.24 rounded to one significant figure is 5
43.261 rounded to one significant figure is 40.
7.3458 rounded to one significant figure is 7.
5 + 40 + 7 is 52 g

Hope this helps you.

Explanation:

7 0

3 years ago

Problem:

kenny6666 [7]

Answer:

a) x = 1.5 *10⁻⁴cos(524πt) m

b) v = -1.5 *10⁻⁴(524π)sin(524πt) m/s

a = -1.5 *10⁻⁴(524π)²cos(524πt) m/s²

c) x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

Explanation:

x = Acos(ωt)

ω = 2πf = 2π(262) = 524π rad/s

x = 1.5 *10⁻⁴cos(524πt)

v = y' = -Aωsin(ωt)

v = -1.5 *10⁻⁴(524π)sin(524πt)

a = v' = -Aω²cos(ωt)

a = -1.5 *10⁻⁴(524π)²cos(524πt)

not sure about the last part as time is generally not given in mm

I will show at 1 second and at 0.001 s to try to cover bases

x(1) = 1.5 *10⁻⁴cos(524π(1))

x(1) = 1.5 *10⁻⁴cos(524π)

x(1) = 1.5 *10⁻⁴(1)

x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = 1.5 *10⁻⁴cos(524π(0.001))

x(0.001) = 1.5 *10⁻⁴cos(0.524π)

x(0.001) = 1.5 *10⁻⁴(-0.0753268)

x(0.001) = -1.129902...*10⁻⁵ m

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

7 0

3 years ago

How much force does it take to bring a 1,050 N car from rest to a velocity of 42 m/s in 13 seconds?

frozen [14]

Answer:

F = 339.23 N

Explanation:

Weight of a car, W = 1050 N

Initial velocity, u = 0

Final velocity, v = 42 m/s

Time, t = 13 s

The weight of an object is given by :

W = mg

g is the acceleration due to gravity

$m=\dfrac{W}{g}\\\\m=\dfrac{1050}{10}\\\\m=105\ kg$

The force acting on car is :

$F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{105\times (42-0)}{13}\\\\F=339.23\ N$

So, the force acting on the car is 339.23 N.

5 0

3 years ago

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)