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Anni [7]
3 years ago
9

Give a graph of velocity v, time does a horizontal line represent ?​

Physics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

The graph represents an object moving at constant velocity

Explanation:

A graph of velocity (v) versus time (t) of an object is generally used to give information about the motion of an object. In fact, from this graph, it is possible to infer several pieces of information:

  • The slope of the graph corresponds to the acceleration of the object. In fact, the acceleration of an object is defined as the rate of change of the velocity: a=\frac{\Delta v}{\Delta t}, which also corresponds to the slope of the graph
  • The area under the curve between two times t_1, t_2 corresponds to the distance travelled by the object in that time interval.

In this problem, we have an object whose velocity-time graph corresponds to a horizontal line. We said that the slope of the graph corresponds to the acceleration of the object: since the slope of an horizontal line is zero, this means that the acceleration of this object is zero, and therefore the object is travelling at constant velocity.

Learn more about velocity:

brainly.com/question/8893949

brainly.com/question/5063905

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the scores of players on a golf team are shown in the table. the teams combined score was 0 what was travis's score?
Alona [7]

Answer:

what table?

Explanation:

3 0
3 years ago
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You want to move a 4- kg bookcase to a different place in the living room. If u push with a force of 65 n and the bookcase accel
IrinaK [193]

Answer:

1.65

Explanation:

The equation of the forces along the horizontal direction is:

F-F_f = ma (1)

where

F = 65 N is the force applied with the push

F_f is the frictional force

m = 4 kg is the mass

a=0.12 m/s^2 is the acceleration

The force of friction can be written as F_f = \mu R (2), where

\mu is the coefficient of kinetic friction

R is the normal force exerted by the floor

The equation of forces along the vertical direction is

R-mg=0 (3)

since the bookcase is in equilibrium. Substituting (2) and (3) into (1), we find

F-\mu mg = ma

And solving for \mu,

\mu = \frac{F-ma}{mg}=\frac{65-(4)(0.12)}{4(9.8)}=1.65

7 0
3 years ago
1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (H
JulijaS [17]

Answer:

The angle between the electric field lines and the equipotential surface is 90 degree.

Explanation:

The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.

The electric field lines are always perpendicular to the equipotential surface.

As

dV = \overrightarrow{E} . d\overrightarrow{r}\\\\

For equipotential surface, dV = 0 so

0 = \overrightarrow{E} . d\overrightarrow{r}\\\\

The dot product of two non zero vectors is zero, if they are perpendicular to each other.

5 0
3 years ago
61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
Mademuasel [1]

Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

P = P_1 + P_2 + P_3 + P_4

here we have

P_1 = (110)(3) = 330 W

P_2 = 100 W

P_3 = 60 W

P_4 = 3 W

P = 330 + 100 + 60 + 3

P = 493 W

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

110\times i = 493

i = 4.48 A

now resistance of transmission line

R = \frac{\rho L}{A}

R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

R = 5.23 \ohm

now power loss in line is given as

P = i^2 R

P = (4.48)^2(5.23)

P = 105 W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{105}{493} \times 100

percentage = 21.3%

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

110 \times 10^3 (i ) = 493

i = 4.48\times 10^{-3} A

now power loss in line is given as

P = i^2 R

P = (4.48 \times 10^{-3})^2(5.23)

P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

percentage = 2.13 \times 10^{-5}%

6 0
3 years ago
A food department is kept at â12°c by a refrigerator in an environment at 30°c. the total heat gain to the food department is
slava [35]

As per energy conservation in the reversible engine we can say

Q_2 + W = Q_1

here we know that

Q_2 = 3300 kJ/h

Q_1 = 4800 kJ/h

now from above equation

3300 + W = 4800

W = 1500 kJ/h

now we can convert it into kW

W = 1500\times \frac{kJ}{3600s}

W = 0.42 kW

so above is the power input to the refrigerator

now to find COP we know that

COP = \frac{Q_2}{W}

COP = \frac{3300}{1500} = 2.2

so COP of refrigerator is 2.2

3 0
3 years ago
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