All categories

3 years ago

A book weighing 5 N rests on top of a table. 1) A downward force of magnitude 5 N is exerted on the book by the force of

Physics

1 answer:

andreev551 [17]3 years ago

7 0

Answer:

1) A downward force of magnitude 5 N is exerted on the book by the force of of gravity

2) An upward force of magnitude 5 N is exerted on the book by the table

Explanation:

First of all, any object near the Earth's surface experiences the forces of gravity, which is also called weight of the object. This force always acts downward.

For the book in the problem, the magnitude of the weight is 5 N.

We also know that the book is at rest: this means that the net force acting on it is zero, and there must be another force balancing the weight, in order to give a zero net force. This other force is the reaction force exerted by the table on the book: the magnitude of this force must be equal to the force of gravity (so, 5 N) and its direction is opposite to the weight, therefore upward.

You might be interested in

Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.

Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

We can solve this first part of the problem by applying the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

$f=\frac{R}{2}$

For this mirror, R = 20 cm, so its focal length is

$f=\frac{20}{2}=+10 cm$ (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm$

We can solve this part by using the magnification equation:

$M=-\frac{y'}{y}=\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

$y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm$

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0

3 years ago

Please correct answers only, if you don’t know, ignore. Please answer this or am I correct?

Wewaii [24]

Answer:

C is correct.

Explanation:

5 0

3 years ago

Read 2 more answers

A honeybee with a mass of 0.150 g lands on one end of a floating 4.75-g popsicle stick, as shown in (Figure 1) . After sitting a

Alekssandra [29.7K]

Answer:

0.0443 m/s

Explanation:

$m_1$ = Mass of honeybee = 0.15 g

$m_2$ = Mass of popsicle stick = 4.75 g

$v_1$ = Velocity of honeybee

$v_2$ = Velocity of stick = 0.14 cm/s

In this system the linear momentum is conserved

$m_1v_1=m_2v_2\\\Rightarrow v_1=\dfrac{m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{4.75\times 0.14}{0.15}\\\Rightarrow v_1=4.43\ m/s$

The velocity of the bee is 4.43 cm/s or 0.0443 m/s

5 0

2 years ago

How much heat is needed to bring 12.0 g of water from 28.3 °C to 43.87 °C, if the specific heat capacity of water is 4.184 J/(g•

Vika [28.1K]

Answer:

Heat capacity, Q = 781.74 Joules

Explanation:

Given the following data;

Mass = 12g

Initial temperature = 28.3°C

Final temperature = 43.87°C

Specific heat capacity of water = 4.184J/g°C

To find the quantity of heat needed?

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 43.87 - 28.3

dt = 15.57°C

Substituting into the equation, we have;

$Q = 12*4.184*15.57$

Q = 781.74 Joules

7 0

2 years ago

State the state the first law of motion

SCORPION-xisa [38]

An object at rest stays at rest and an object in motion<span> stays in </span>motion <span>with the same speed and in the same direction unless acted upon by an unbalanced force.
I hope this helps you </span>

3 0

3 years ago