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2 years ago

Question 10 of 34

1 answer:

3 0

Julia walks from the park, which is six blocks east of her house, to the store, which is three blocks east of her house. Julia walks for 5 minutes. This walk's average speed will be 1.2 blocks per minute. Option B is correct.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points. Distance is the horizontal length covered by the body. While displacement is the shortest distance between the two points.

Displacement is a vector quantity .its unit is m.

The average velocity on this walk will be;

$\rm v_{avg}= \frac{d}{t} \\\\ \rm v_{avg}= \frac{6 \ block+ 3 \ block }{5 \ minute } \\\\ v_{avg}=1.4 \ block /min$

Hence option B is correct.

To learn more about displacement refer to the link; brainly.com/question/10919017

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The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c

Natalka [10]

Answer:

$EMF = 1684.67 Volts$

Explanation:

As we know that EMF is induced in a closed conducting loop if the flux linked with the loop is changing with time

So we can say

$EMF = \frac{d\phi}{dt}$

now we have

$\phi = BA$

here since magnetic field is constant so we have

$EMF = A\frac{dB}{dt}$

now we have

$A = (190 \times 10^3)(190 \times 10^3)$

$A = 3.61 \times 10^{10} m^2$

now we have

$EMF = 3.61\times 10^{10} (\frac{2.8 \times 10^{-6}T}{60 s})$

$EMF = 1684.67 Volts$

6 0

3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Vlad [161]

Answer: $7.95^{\circ}C$

Explanation:

Given

Mass of water is $m_1=1\ kg$

mass of ice is $m_2=0.1\ kg$

Latent heat of fusion $L=3.33\times 10^5\ kJ/kg$

The heat capacity of water is $c_{w}=4186\ J/kg^{\circ}C$

Suppose water is at $T^{\circ} C$ and it reaches to $0^{\circ}C$ to melt the ice

the heat released by water must be equivalent to heat absorbed by the ice

$\therefore \quad m_1c_w(T-0)=m_2\times L\\\Rightarrow 1\times 4186\times T=0.1\times 3.33\times 10^5\\\Rightarrow T=7.95^{\circ}C$

6 0

2 years ago

A frog hops 5m east and 2m north. What is the magnitude of the frogs total displacement in m?

TiliK225 [7]

As per the question a frog jumps 5 m towards east.

Frog again jumps 2 m north.

Let the displacement along east is denoted by vector A and the displacement towards north is denoted as vector B.

Hence magnitude of A = 5 m

Magnitude of B = 2 m

We are asked to calculate the total displacement.

Here the angle between them is 90 degree as A is towards east and B is towards north.

As per parallelogram law of vector addition,the magnitude of total displacement [R] will be-

$R=\sqrt{ A^{2} +B^{2}+2AB\cos\theta}$

$=\sqrt{ 5^{2}+ 2^{2}+2*5*2 cos 90}$

$=\sqrt{25+4+0} m$ [cos90= 0]

$=\sqrt{29} m$

$= 5.38516 m$ [ans]

7 0

3 years ago

To see how deep a well is you drop a rock in. The rock hits the water 10 seconds later.

Karolina [17]

Answer:

490m

Explanation:

Given parameters:

Time taken = 10s

Unknown:

Depth of the well = ?

Solution:

To find the depth of the well, we apply the appropriate motion equation:

h = ut + $\frac{1}{2}$ gt²

h is the depth

u is the initial velocity

g is the acceleration due to gravity = 9.8m/s²

t is the time taken

Input the parameters and solve;

The ball was dropped from rest; u = 0

h = $\frac{1}{2}$ gt²

h = $\frac{1}{2}$ x 9.8 x 10² = 490m

6 0

3 years ago

A hose has been clamped so that the area at the clamp is only one quarter the area of the rest of the hose. When we ignore the v

Bad White [126]

Answer:

Explanation:

A hose has been clamped so that the area at the clamp is only one quarter the area of the rest of the hose. When we ignore the viscosity of water, the ratio of the volume of water delivered per unit time when the clamp is on to the volume of water delivered per unit time without the clamp is 1 as continuity says the same amount of water must flow out

4 0

3 years ago