Which best explains why the diagram shows refraction but not reflection?

Katie and Mark sit next to one another in class. She has a mass of 40 kg and his mass is 65 kg.

OlgaM077 [116]

A :-) for this question , we should apply
F = GMm by d^2
( For making the calculation easy , first remove the decimals )
Given : G = 6.7 x 10^-11 Nm^2 / kg^2
= 67 x 10^-12 Nm^2 / Kg^2
M = 65 kg
m = 40 kg
d = 0.5 m
Solution -
F = GMm by d^2
F = 67 x 10^-12 x 65 x 40 by 0.5 x 0.5
F = 4355 x 40 x 10^-12 by 0.25
F = 174200 x 10^-12 by 0.25
F = 696800 x 10^-12

.:. The Gravitational force between mark and Katie is 696800 x 10^-12

6 0

3 years ago

An object of mass 2kg is on an incline where there is an applied force of 15N. The

seraphim [82]

a) See free-body diagram in attachment

b) The acceleration is $2.46 m/s^2$

Explanation:

a)

The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.

The free-body diagram for this object is shown in the figure in attachment.

There are three forces acting on the object:

The weight of the object, labelled as $mg$ (where m is the mass of the object and g is the acceleration of gravity), acting downward
The applied force, $F_a$ , acting up along the plane
The force of friction, $F_f$ , acting down along the plane

b)

In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:

$F_a - F_f - mg sin \theta = ma$

where:

$F_a = 15 N$ is the applied force, pushing forward

$F_f = 5 N$ is the frictional force, acting backward

$mg sin \theta$ is the component of the weight parallel to the incline, acting backward, where

m = 2 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=15^{\circ}$ is the angle between the horizontal and the incline (it is not given in the problem, so I assumed this value)

a is the acceleration

Solving for a, we find:

$a=\frac{F_a - F_f - mg sin \theta}{m}=\frac{15-5-(2)(9.8)(sin 15^{\circ})}{2}=2.46 m/s^2$

Learn more about inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

8 0

3 years ago

A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 10 A, and that o

Katyanochek1 [597]

Answer:

Part a)

P = 1200 Watt

Part b)

P = 480 W

Part c)

$\frac{E_{bd}}{E_{vc}} = 5 : 2$

Explanation:

Part a)

Power consumed by blow dryer is given as

$P = iV$

here supply voltage is given as

$V = 120 volts$

current rating of blow dryer is given as

$i = 10 A$

$P = (120 V)(10 A)$

$P = 1200 W$

Part b)

Power consumed by vacuum cleaner is given as

$P = iV$

here supply voltage is given as

$V = 120 volts$

current rating of vacuum cleaner is given as

$i = 4 A$

$P = (120 V)(4 A)$

$P = 480 W$

Part c)

Energy consumed in a given interval of time is given as

$Energy = power \times time$

now we have to find the ratio of energy consumed in time interval of 11 minutes

so we have

$\frac{E_{bd}}{E_{vc}} = \frac{P_{bd} t}{P_{vc} t}$

now we have

$\frac{E_{bd}}{E_{vc}} = \frac{1200 W\times 11 min}{480 W \times 11 min}$

$\frac{E_{bd}}{E_{vc}} = 5 : 2$

5 0

4 years ago

Mr. Smith challenged his students to create an electromagnet that will pick up more paper clips than anyone else in the class. T

aliya0001 [1]

Wrap the wire very tight around the iron nail.

5 0

3 years ago

Read 2 more answers

(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind

Vilka [71]

There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
$F_f = \mu N$
where $\mu$ is the coefficient of friction, while N is the normal force. So we have:
$F=\mu N$
since we know that F=300 N and $\mu=0.3$ , we can find N, the magnitude of the normal force:
$N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N$

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is $\mu=0.03$ due to the presence of the oil. Therefore, we have:
$N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N$

8 0

3 years ago