Help asap please!

One end of an elastic cord is fastened to a steel beam. A metal weight with a mass of 68 kg is attached to the other end of the

Vika [28.1K]

Hi there!

We can use IMPULSE to solve.

Recall impulse:

$I = \Delta p = m\Delta v = m(v_f - v_i)$

Begin by calculating the impulse. Assuming up to be the + direction, and down to be the - direction.

$I = 68(11 - (-14)) = 68 \cdot 25 = 1700 kg\frac{m}{s}$

Now, we can calculate force using this value:

$I = F \cdot t\\\\F = \frac{I}{t}\\\\F = \frac{1700}{0.85} = \boxed{2000 N}$

<u>The weight experiences a net force of 2000N UPWARDS.</u>

4 0

3 years ago

For what value of the ratio r/a of plate radius to separation between the plates does the electric field at the point x=a/2 on t

SIZIF [17.4K]

solution:

E\delta =\frac{R}{\epsilon0}(1-\frac{A}{\sqrt{4R^{2}}-ac}

=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4r^{2}/^{_a{2}}+1}})

=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4x^2+1}})

x=\frac{r}{a}

infinite case,

Ei=\frac{r}{\epsilon0}

\therefore e\delta =ei(1-\frac{1}{\sqrt{4x^{2}+1}})

we have to find x when,

ei-e\delta =1% ,y=ei=1/100 ei

or,ei-ei+\frac{ei}{\sqrt{4x^2+1}} = 1/100ei

\frac{1}{\sqrt{4x^2+1}}=\frac{1}{100}

4x^2+1 =10^4

x=\frac{\sqrt{\frac{10^4-1}{4}}}=49.99\approx 50

\therefore \frac{r}{a}\approx 50

6 0

3 years ago

The weight of a 72.0 kg astronaut on the Moon, where g = 1.63 m/s2 is (5 points) Select one: a. 112 N b. 117 N c. 135 N d. 156 N

Hunter-Best [27]

Answer: The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

Explanation:

Mass of the astronaut on the moon , m= 72 kg

Acceleration due to gravity on moon,g = 1.63 $m/s^2$

According to Newton second law of motion: F = ma

This will changes to = Weight = mass × g

$Weight=72 kg\times 1.63m/s^2=117.36 N$

The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

5 0

4 years ago

Read 2 more answers

Critical thinking!!

Maslowich

Answer:

No there is no displacement observed so no work done

8 0

4 years ago

A speed truck locks it’s brakes and it skids to a stop. If the truck’s total mass were doubled, it’s skidding distance would be?

Gnom [1K]

Answer:

The skidding distance would be doubled

Explanation:

When the truck applies the brakes and slows down, its motion is a uniformly accelerated motion, so its skidding distance can be found by using the suvat equation

$v^2-u^2=2as$

where

v = 0 is the final velocity (zero since the truck comes to a stop)

u is the initial velocity

a is the acceleration

s is the skidding distance

The acceleration can also be written as

$a=\frac{F}{m}$

where F is the force applied by the brakes and m the mass of the truck. Substituting into the previous equation,

$-u^2 = 2\frac{F}{m}s\\s = -\frac{mu^2}{2F}$

We see that the skidding distance is proportional to the mass: therefore, if the mass of the truck is doubled, the skidding distance will double as well.

5 0

3 years ago