Does anyone know the answer to this page and the next one that it has?

A woman exerts a constant horizontal force on a large box. As a result, the box moves across a horizontal floor at a constant sp

d1i1m1o1n [39]

Answer:C

Explanation:

When a constant horizontal force is applied to the box, box started moving in the horizontal direction such that it moves with constant velocity $v_0$

Constant velocity implies that net force on the box is zero

i.e. there must be an opposing force which is equal to the applied force and friction force can serve that purpose.

So option c is the correct choice.

6 0

4 years ago

An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Alexeev081 [22]

Answer:

a) $c=1822.3214\ J.kg^{-1}.K^{-1}$

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

mass of aluminium, $m_a=0.1\ kg$

mass of water, $m_w=0.25\ kg$

initial temperature of the system, $T_i=10^{\circ}C$

mass of copper block, $m_c=0.1\ kg$

temperature of copper block, $T_c=50^{\circ}C$

mass of the other block, $m=0.07\ kg$

temperature of the other block, $T=100^{\circ}C$

final equilibrium temperature, $T_f=20^{\circ}C$

We have,

specific heat of aluminium, $c_a=910\ J.kg^{-1}.K^{-1}$

specific heat of copper, $c_c=390\ J.kg^{-1}.K^{-1}$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

$Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)$

$Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)$

$Q_{in}=11375\ J$

The heat released by the blocks when dipped into water:

$Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)$

where

$c=$ specific heat of the unknown material

For the conservation of energy : $Q_{in}=Q_{out}$

so,

$11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)$

$c=1822.3214\ J.kg^{-1}.K^{-1}$

b)

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

7 0

3 years ago

Intelligence tests that are given to participants in groups

White raven [17]

a. are typically paper-and-pencil measures.

Similar with psychological tests, mostly structured personality tests.

Psychological tests comes two ways: 
The structure psychological tests or, objectives tests and unstructured psychological tests or, also called projective tests. By what you are referring the responder strongly asserts a projective tests which in definition comes with an unambiguous stimuli or no paper test just drawings and images. If what the responder’s suggesting is correct you are referring to the Rorschach projective tests, these tests are a figure symmetrically placed in an inkblot that lets you visualize or create a mental picture out of it, and makes you describe what you in see much detail as you can.

4 0

3 years ago

A ship, carrying freshwater to a desert island in the caribbean, has a horizontal cross-sectional area of 2800 m2 at the waterli

Alex17521 [72]

Due that we already know the horizontal cross-sectional area of the ship, which is 2800 m2 and we are going to understand that value keeps constant for the whole 9.5 of height of the ship from the waterline till the new waterline after unloading, then we just need to calculate the volume as follows: V = A * H , where V is volume, A is area and H is height V= 2,800 * 9.5 = 26,600 m3 So this volum of 26,600 cubic meters is the volum of freshwater delivered in the island.

3 0

3 years ago

Suppose you have a 113-kg wooden crate resting on a wood floor. (μk = 0.3 and μs = 0.5) (a) What maximum force (in N) can you ex

AURORKA [14]

Answer:

554.27N

Explanation:

(a) The max frictional force exerted horizontally on the crate and the floor is,

Substitute the values,

μs=0.5

mass=113kg

g=9.81m/s

Ff=μsN

=μsmg

=(0.5 x 113 x 9.81)

Ff=554.27N

3 0

3 years ago