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Paha777 [63]
3 years ago
9

A student on her way to school walks eastward in a straight line 20.0 meters towards the bus stop, but realizes she dropped her

physics notebook along the way and turns around to retrace her steps. She has traveled back 18.0 meters before finding it and then runs the remaining 45.0 meters due east to the bus stop. What is her total displacement? What is her total distance traveled?
Physics
1 answer:
larisa [96]3 years ago
3 0

Answer:

Total displacement will be 47 meter

Total distance will be 83 meters

Explanation:

We have given that first the student go eastward towards bus stop 20 meters

But he realizes that she dropped his physics notebook and so h=she turns back along the same way up to 18 meters

So displacement = 20-18 = 2 meters

And he travel 45 meters in east along the bus stop so total displacement = 45+2 = 47 meters

Total distance traveled by the student = 20+18+45 = 83 meters  

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You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh
lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

F=\frac{kQq}{r^2}

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 45^\circ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

#SPJ4

3 0
2 years ago
You hang a heavy ball with a mass of 30 kg from a tungsten rod 2.8 m long by 1.5 mm by 2.6 mm. You measure the stretch of the ro
guajiro [1.7K]

Answer:

Young's modulus (Y) = 3.56×10^11 N/m^2

The speed of sound in tungsten = 6166.4 m/s

Explanation:

Young's modulus (Y) = stress/strain

Stress = force/area

Force = mg = 30×9.8 = 294 N

Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2

Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2

Strain = extension/length

Extension = 0.000594 m

Length = 2.8 m

Strain = 0.000594/2.8 = 2.12×10^-4

Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2

Y = h × rho × g

rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3

h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m

From the equations of motion

v^2 = u^2 + 2gh =

Initial speed (u) = 0 m/s

v = sqrt (2×9.8×1.94×10^6)

v = 6166.4 m/s

7 0
3 years ago
Read 2 more answers
You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand at 30 m/s. What is the kinetic energy (KE) of the volley
FinnZ [79.3K]

awnser :

Explanation:

8 0
2 years ago
Read 2 more answers
A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125
frez [133]

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

7 0
3 years ago
will someone give me 2 brainleist PLZZZZZZZZ andif you do i will Give 20 ponits and i will be you friend
Zanzabum
How do you want me to give you points tell me and I’ll do it
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2 years ago
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