96 K = _____<br> -177°C<br> 0°C<br> 96°C<br> 369°C

How much force was applied to an object if was moved 2 meters and the work done on the object was 8 joules?

serious [3.7K]

Answer:

The answer is 4N or B

Explanation:

Just the equation W = F x D.

We have W = 8 J and D = 2 m using algebra ....

8J/2m = F ... F = 4.

6 0

3 years ago

Read 2 more answers

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t , where the time tis in seconds. The

Reika [66]

Complete question:

The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The

particle is momentarily at rest at t is:

Select one:

a. 9.3s

b. 1.3s

C. 0.75s

d.5.3s

e. 7.3s

Answer:

b. 1.3 s

Explanation:

Given;

position of the particle, x(t)=1 6t- 3.0t³

when the particle is at rest, the velocity is zero.

velocity = dx/dt

dx /dt = 16 - 9t²

16 - 9t² = 0

9t² = 16

t² = 16 /9

t = √(16 / 9)

t = 4/3

t = 1.3 s

Therefore, the particle is momentarily at rest at t = 1.3 s

6 0

3 years ago

En un m.A.S. La amplitud tiene un valor de 10 centimetros y el periodo es de 2 segundos calcular el valor de la velocidad de 0.8

Ivanshal [37]

Answer:

v1=18.46m/s

v2=29.8cm/s

Explanation:

We know that

$A=10cm\\T=2s$

the equation of the motion is

$x=Acos(\omega t)\\$

we can calculate w by using

$\omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi$

Hence, we have that

$x=10cm*cos(\pi t)\\$

the speed will be

$v=-\omega*Asin(\omega t)\\|v(0.8)|=|\pi*10cm*sin(\pi *0.8)|=18.46\frac{cm}{s}\\|v(1.4)|=|\pi*10cm*sin(\pi *1.4)|=29.8\frac{cm}{s}$

hope this helps!

6 0

3 years ago

Convert 5.5 kilometers into millimeters.

dimaraw [331]

Answer:

5500000 millimeters

Explanation:

1 kilometre= 1000 meter

5.5 km=5.5 * 1000

=5500

Now,

1 metre = 1000 millimetres

5500 metre=1000*5500

=5500000 mm

4 0

1 year ago

We need to find the launch velocity of our new marble launcher. we know that it will launch a 25g marble to a distance of 73 cm,

White raven [17]

The launch velocity of the marble launcher is 34.65 m/s

Given that the launch velocity of marble launcher, launches a 25g marble to a distance of 73 cm (0.73 m) and the marble roll up to 6.2 meters before stopping. The launch height is 20 cm (0.2 m).

The time for landing can be calculated by the second equation of motion formula:

h = ut + $\frac{1}{2}$ g $t^{2}$

Let u = 0

0.2 = 0×t + $\frac{1}{2}$ × 9.8 × $t^{2}$

$t^{2}$ = $\frac{0.2}{4.9}$

$t^{2}$ = 0.04

t = 0.2s

Now, the launch velocity of the marble launcher can be calculated by:

Speed = Distance / Time

Speed = $\frac{0.73 + 6.3}{0.2}$

Speed = $\frac{6.93}{0.2}$

Speed = 34.65 m/s

Therefore, the launch velocity of the marble launcher is 34.65 m/s

Know more about Launch velocity: -brainly.com/question/18883779

#SPJ9

3 0

1 year ago

96 K = _____ -177°C 0°C 96°C 369°C