All categories

2 years ago

DUE BY MIDNIGHT

Physics

1 answer:

olga2289 [7]2 years ago

8 0

Answer:

Option D. 1000 J.

Explanation:

From the question given above, the following data were obtained:

Force (F) applied = 200 N

Distance (s) = 5 m

Time (t) = 10 s

Workdone (Wd) =?

Workdone (Wd) is simply defined as the product of force (F) and distance (s) moved in the direction of the force. Mathematically, it is expressed as:

Wd = F × s

With the above formula, we can calculate the Workdone as illustrated below:

Force (F) applied = 200 N

Distance (s) = 5 m

Workdone (Wd) =?

Wd = F × s

Wd = 200 × 5

Wd = 1000 J

Thus, the Workdone is 1000 J

You might be interested in

A scientific theory _______.

Nesterboy [21]

B. is not a validated bu experimentation

5 0

2 years ago

Read 2 more answers

You are riding on a Ferris wheel that is rotating with constant speed. The car in which you are riding always maintains its corr

nirvana33 [79]

It's going down.

Hope this helps you

5 0

3 years ago

A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continu

kolbaska11 [484]

Answer:

τ=0.060 N.m

Explanation:

By kinematics:

$\omega f = \omega o-\alpha*t$

Solving for α:

$\alpha=\frac{\omega o-\omega f}{t}$

where ωo = 600*2*π/60; ωf = 0; t=10s

$\alpha=6.283rad/s^2$

The sum of torque is:

$\tau=I*\alpha$

$\tau=M*R^2/2*\alpha$

$\tau=0.060 N.m$

8 0

3 years ago

Flammable materials, like alcohol, should never be dispensed or used near

Vaselesa [24]

Answer:an open flame

Explanation:

Flammable materials like alcohol should never be dispensed or used near an open flame

3 0

2 years ago

Read 2 more answers

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of

garri49 [273]

Answer:

$w_{0}=14$

$t=\frac{\pi }{14}$

Explanation:

Data

mass m= 100g

Length L= 5cm

we can use:

gm-kL= 0

divide both side by m

g - $\frac{kL}{m}$ =0

where

$\frac{k}{m}$ = $\frac{g}{L}$

$\frac{k}{m}=w_{0}$ ^{2}

so now

$w_{0}^{2}$ = $\frac{9.8*100}{5}$

$w_{0}^{2}=\frac{980}{5}$

$w_{0}^{2}=196$

square both side

$w_{0}=\sqrt{196}$

$w_{0}=14$

We can apply:

u(t)=Acoswt +Bsinwt

u(t)=Acos14t +Bsin14t

u(0)=0 where A=0

therefore

u(0) = Bsin14t

$u^{'}$ (0) = 10 ⇒ 10=14B ⇒ B= $\frac{14}{10}$ B= $\frac{5}{7}$

so now u(t)= $\frac{5}{7}$ sin14t

so t will be:

t= $\frac{\pi }{14}$

t= $\frac{3.14}{14}$

t=0.22 seconds

8 0

3 years ago