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3 years ago

15

Cindy exerts a force of 40 N and moves a chair 6 m. Her brother and he pushes a different chair for 6 m while exerting a force o

f 30 N. Andy does: more, less work than Cindy. The difference of their amounts of work is: 20, 40, 60, or 180 Nm.

1 answer:

Lostsunrise [7]3 years ago

6 0

Wa = F*d = 240J
Wb = 180 J

Andy does less work
Difference is 60J

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A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m

Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

$K.E+P.E=P.E+K.E$

$\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2$

$\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}$

Put the value into the formula

$\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}$

$\dfrac{v_{1}^2}{2}=18.62$

$v_{1}=\sqrt{2\times18.62}$

$v_{1}=6.10\ m/s$

Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec

MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

$\frac{1}{2} mu_1^2+0=\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2$

∴ $\frac{1}{2} m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2$

∴ $\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2$

∴ $mv$ ₁ $v$ ₂ = 0

It is impossible for the mass to be zero.
Because the second ball moves, velocity v2 cannot be zero.
As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.

<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

#SPJ4

3 0

1 year ago

What displacement in cm would occur with a 75 N/m spring if you placed a 300 N weight on the spring?

ollegr [7]

Surface tension=75N/m

Weight=300N

$\\ \bull\tt\longmapsto Surface\:Tension=\dfrac{Weight}{Displacement}$

$\\ \bull\tt\longmapsto Displacement=\dfrac{Weight}{Surface\:Tension}$

$\\ \bull\tt\longmapsto Displacement=\dfrac{300}{75}$

$\\ \bull\tt\longmapsto Displacement=4m$

$\\ \bull\tt\longmapsto Displacement=400cm$

6 0

2 years ago

A plane is landing. It started at 400m/s and ended up at 50m/s after 30 seconds. What is it's acceleration?

quester [9]

Give u = start velocity

v = end velocity

v = u + at

50 = 400 + a*30

30a = -350

a = -116.67 m/ $s^{2}$

**Why the accecleration is negative number**

Because displacement, velocity, and acceleration are VECTOR QUANTITIES.

Vector Quantity must have direction.

4 0

3 years ago