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lisabon 2012 [21]

2 years ago

9

Four students are each pushing on one side of a box. Mara is pushing to the west, Cindi is pushing to the east, Chris is pushing

to the north, and Sammy is pushing to the south. If the box moves to the east, which student does work on the box?

1 answer:

jenyasd209 [6]2 years ago

5 0

Answer:

cindi

Explanation:

cindi's work done is larger than all the other students combined

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A 92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. The coefficient of k

9966 [12]

Answer:

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg : mass of the skier

θ =30° :angle θ of the ski slope with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W= 92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N= 0.10*780.8 N

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax , ax= a : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 = (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²

6 0

2 years ago

A rabbit is moving in the positive x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 10.9 m/s al

anzhelika [568]

Answer:

aₓ = 0 , ay = -6.8125 m / s²

Explanation:

This is an exercise that we can solve with kinematics equations.

Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.

x axis

vₓ = v₀ₓ = 1.10 m / s

aₓ = 0

y axis

initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

$v_{y}$ = v_{oy} -ay t

ay = (v_{oy} -v_{y}) / t

ay = (0 -10.9) / 1.6

ay = -6.8125 m / s²

the sign indicates that the acceleration goes in the negative direction of the y axis

8 0

2 years ago

A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu

Vesnalui [34]

Answer:

The magnitude of the resultant acceleration is 2.2 $m/s^2$

Explanation:

Mass (m) of the sailboat = 2000 kg

Force acting on the sailboat due to ocean tide is $F_1$ = 3000N

Eastwards means takes place along the positive x direction

Then $F_{1x}$ = 3000N and $F_{1y}$ = 0

Wind Force acting on the Sailboat is $F_2$ = 6000N directed towards the northwest that means at an angle 45 degree above the negative x axis

Then

$F_{2x}$ = -(6000N) cos 45 degree = -4242.6 N

$F_{2y}$ = (6000N) cos 45 degree = 4242.6 N

Hence , the net force acting on the sailboat in x direction is

$F_x = F_{1x}+ F_{2x}$

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

Net Force acting on the sailboat in y direction is

$F_y = F_{1y}+ F_{2y}$

= 0+ 4242.6N

= 4242.6N

The magnitude of the resultant force =

Using pythagorean theorm of 1243 N and 4243 N

$\sqrt{(1242.6)^2 + (4242.6)^2$

$\sqrt{(1544054.76) + (17999654.8)}$

$\sqrt{(19543709.5)^2}$

4420.8 N

F = ma

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

$a =\frac{4420.8}{ 2000}$

=2.2 $m/s^2$

4 0

2 years ago

A tennis player standing 12.6m from the net hits the ball at 3.00 degrees above the horizontal. To clear the net, the ball must

mezya [45]

We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)

4 0

2 years ago

Read 2 more answers

Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual

kotegsom [21]

Answer:

The amount of heat required is $H_t = 1.37 *10^{6} \ J$

Explanation:

From the question we are told that

The mass of water is $m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g$

The temperature of the water before drinking is $T_w = 3.8 ^oC$

The temperature of the body is $T_b = 36.6^oC$

Generally the amount of heat required to move the water from its former temperature to the body temperature is

$H= m_w * c_w * \Delta T$

Here $c_w$ is the specific heat of water with value $c_w = 4.18 J/g^oC$

So

$H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)$

=> $H= 7.8 *10^{4} \ J$

Generally the no of mole of sweat present mass of water is

$n = \frac{m_w}{Z_s}$

Here $Z_w$ is the molar mass of sweat with value

$Z_w = 18.015 g/mol$

=> $n = \frac{5.7 *10^2}{18.015}$

=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

=> $H_v = 31.6 * 7 *10^{3}$

=> $H_v = 1.29 *10^{6} \ J$

Generally the overall amount of heat energy required is

$H_t = H + H_v$

=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

4 0

2 years ago