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3 years ago

5

A commuting student leaves home and drives to school at an average speed of 39.0 km/h. After 23.0 min he realizes that he has fo

rgotten his homework and returns home to get it at the same average speed. It takes 10.0 min to find the report, after which the trip to school 39.0 km away to the east is resumed at the same speed as before. What is the average speed for the entire trip ? What is the average velocity for the entire trip?

1 answer:

Ilia_Sergeevich [38]3 years ago

8 0

Answer: $s_{avg}=\frac{117}{\frac{213}{60}}=32.95 km/h$

$v_{avg}=\frac{39}{\frac{213}{60}}=10.98 km/h$

Explanation:

Given

Distance between home and school is 39 km

Total time taken $=\frac{39}{39}+\frac{23}{60}+\frac{39}{39}+\frac{10}{60}+\frac{39}{39}$

i.e.

$t=3\times 60+23+10=213 min$

Total Distance traveled =39+39+39=117 km

displacement=39 km

Thus Average speed $=\frac{Distance}{Time\ taken}$

$s_{avg}=\frac{117}{\frac{213}{60}}=32.95 km/h$

Average velocity $=\frac{Displacemet}{Time\ taken}$

$v_{avg}=\frac{39\times 60}{213}=10.98 km/h$

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The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?

timurjin [86]

Answer: The drag force goes up by a factor of 4

Explanation:

The <u>Drag Force</u> equation is:

$F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2}$ (1)

Where:

$F_{D}$ is the Drag Force

$C_{D}$ is the Drag coefficient, which depends on the material

$\rho$ is the density of the fluid where the bicycle is moving (<u>air in this case) </u>

$A_{D}$ is the transversal area of the body or object

$V$ the bicycle's velocity

Now, if we assume $C_{D}$ , $\rho$ and $A_{D}$ do not change, we can rewrite (1) as:

$F_{D}=C.V^{2}$ (2)

Where $C$ groups all these coefficients.

So, if we have a new velocity $V_{n}$ , which is the double of the former velocity:

$V_{n}=2V$ (3)

Equation (2) is written as:

$F_{D}=C.V_{n}^{2}=C.(2V)^{2}$

$F_{D}=4CV^{2}$ (4)

Comparing (2) and (4) we can conclude<u> the Drag force is four times greater when the speed is doubled.</u>

7 0

3 years ago

The current through a 10 ohm resistor connected to a 120 volt power supply is

Leno4ka [110]

Answer:I=12 A

Explanation:

Given

Resistance $R=10 \Omega$

Voltage $V=120 V$

According to ohm's law current through a conductor is directly proportional to the voltage applied.

$V\propto I$

$V=IR$

where V=Voltage

I=Current

R=resistance

$I=\frac{V}{R}$

$I=\frac{120}{10}$

$I=12 A$

4 0

3 years ago

The distance that a spring will stretch varies directly as the force applied to the spring. A force of 8080 pounds is needed to

xxTIMURxx [149]

Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x Formula (1)

Where;

F is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21 inches

Problem development

We replace data in formula 1 to calculate K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate F₂

F₂= K*x₂

F₂= (10)*(21)

F₂= 210 pounds

8 0

3 years ago

monitta

Answer:

See explanation below.

Explanation:

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7 0

2 years ago

2. A woman prevents a 3kg brick from falling by pressing it against a vertical wall. The coefficient of friction

Anettt [7]

<h3>Answer:</h3>

49 N

<h3>Explanation:</h3>

<u>We are given;</u>

Mass of the brick as 3 kg
The coefficient of friction as 0.6

We are required to determine the force that must be applied by the woman so the brick does not fall.

We need to importantly note that;
For the brick not to fall the, the force due to gravity is equal to the friction force acting on the brick.

That is; Friction force = Mg

But; Friction force = μ F

Therefore;

μ F = mg

0.6 F = 3 × 9.8

0.6 F = 29.4

F = 49 N

Therefore, she must use a force of 49 N

6 0

3 years ago