Ask question

Ask question

All categories

2 years ago

15

Help with this question plz and show how you got it :) tysm!

1 answer:

spayn [35]2 years ago

5 0

The hardest part of the job is to find the right formula to use, and write it down. You've already done that ! The rest is just turning the crank until an answer falls out.

You wrote. E = m g h.
Beautiful.
Now divide each side by (g h), and you'll have the formula for mass:

m = E / (g h).

You know all the numbers on the right side. Just pluggum in, do the arithmetic, and you'll have the mass.

You might be interested in

Mr. Barcena reached a top speed of 5m/s in 2 seconds from rest, what was his average acceleration including units?

sashaice [31]

Answer:

Explanation:

The equation for acceleration is

$a=\frac{v_f-v_0}{t}$ where vf is the final velocity and v0 is the initial velocity. Filling in:

$a=\frac{5-0}{2}$ so

a = 2.5 m/s/s

4 0

2 years ago

An electron that has moved from the outer energy levels towards the inner energy levels of an atom is said to be in a ground sta

IRISSAK [1]

I think it’s true....

7 0

2 years ago

Force is a vector because it has both ___

scoray [572]

Answer:

C. Size and Direction

6 0

2 years ago

A car travels along a highway with a velocity of 24 m/s, west. The car exits the highway; and 4.0 s later, its instantaneous vel

zloy xaker [14]

Answer:

$4.25 m/s^{2}$

Explanation:

Change in velocity considering the x component will be

Final velocity-Initial velocity

$\triangle v_x= 16cos 45^{\circ}-24=-12.6862915 m/s$

Change in velocity considering the y component will be

Final velocity-Initial velocity

$\triangle v_y= 16sin 45^{\circ}-0=11.3137085 m/s$

Resultant change in velocity $=\sqrt {(-12.6862915 m/s)^{2}+(11.3137085 m/s)^{2}}=16.9982938 m/s$

Acceleration= change in velocity per unit time hence

$a= \frac {16.9982938}{4}=4.24957345\approx 4.25 m/s^{2}$

5 0

3 years ago

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force = $\mu \left ( N\right )=0.4\times \left ( 740+160\right )$

=360 N

$cos\theta =\frac{3}{5}$

$sin\theta =\frac{4}{5}$

(b)Moment about Ground Point

$740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

$N_1tan\theta =1140$

$N_1=171 N$

$N_1=f=171 N$

(c)

$740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

Here maximum friction force can be 360 N

Therefore $N_1=360 N$

Where x is the maximum distance moved by man along the ladder

$360\times 5\times \frac{4}{3}=740x+160\times 2.5$

740x=2000

x=2.702m

5 0

2 years ago