Ask question

Ask question

All categories

3 years ago

15

Please help on this one

1 answer:

mixas84 [53]3 years ago

6 0

Ep=mgh
h= Ep/mg
h=57÷(3.3×9.8)
h= 57÷32.34
h= 1.8m
So; the answer is B. 1.8m

You might be interested in

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

pantera1 [17]

Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$

The sum of the vertical components of F1 and F2 must equal the total force Ft

$\displaystyle -2F\ cos21^o+F\ sin\alpha =460$

Solving for $\alpha$ in the first equation

$\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o$

$\displaystyle cos\alpha =0.717=>\alpha =44.214^o$

$\displaystyle F(2cos21^o+sin\alpha)=460$

$\displaystyle F=\frac{460}{2cos21^o+sin\alpha}$

$\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}$

$\displaystyle F=179.37\ N$

The magnitude of F1 is

$|F1|=2*F=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

4 0

3 years ago

The diagram below represents a light ray striking the surface of a flat, shiny object.

Tanzania [10]

Answer:

<em><u>option</u></em><em><u> (</u></em><em><u>C)</u></em><em><u> </u></em><em><u>is </u></em><em><u>right</u></em><em><u> answer</u></em>

Explanation:

I think it's helps you

5 0

3 years ago

Read 2 more answers

On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around

Mekhanik [1.2K]

Answer:

$v\approx 16.956\,\frac{m}{s}$

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

$\Sigma F = f = m\cdot \frac{v^{2}}{R}$

The maximum speed is:

$v = \sqrt{\frac{f\cdot R}{m} }$

$v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }$

$v\approx 16.956\,\frac{m}{s}$

7 0

3 years ago

A net force of 24 N is acting on a 4.0-kg object. Find the acceleration in m/s2.

Inessa [10]

Hi there!

We can use Newton's Second Law:
$\Sigma F = ma$

ΣF = Net force (N)
m = mass (kg)
a = acceleration (m/s²)

We can rearrange the equation to solve for the acceleration.

$a = \frac{\Sigma F}{m}\\\\a = \frac{24}{4} = \boxed{6 \frac{m}{s^2}}$

6 0

2 years ago

An object on Earth weighs 150 N. What is its mass?

Alex777 [14]

Answer:

$\tt \: mass \: = \frac{weight}{acceleration \: due \: to \: gravity}$

$\longrightarrow \tt \: \frac{150}{10}$

$\longrightarrow \boxed{ \tt{15 \: kg}}$

Our final answer is 15 kg .

----------- HappY LearninG<3 ------------

3 0

2 years ago