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alexandr1967 [171]
3 years ago
11

Tiana jogs 1.5 km along a straight path and then turns and jogs 2.4 km in the opposite direction. She then turns back and jogs 0

.7 km in the original direction. Let Tiana's original direction be the positive direction. What are the displacement and distance she jogged?
Physics
1 answer:
vichka [17]3 years ago
7 0

Answer:

Distance: 4.6km Displacement= -0.2km

Explanation:

Total distance: 1.5+2.4+0.7= 4.6 km

Displacement: 1.5-2.4+0.7= -0.2km

The displacement may also be 0.2km, it just depends on if it wants it negative or not.

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Stars are classified mainly according to their distance from Earth.
Zielflug [23.3K]
 The answer is b) false. Stars are mainly classified by their spectra (the elements they absorb) and their temperature.
4 0
3 years ago
A 0.1ohm resistor has power rating 5w. is the resistor safe when conducting a current of 10A.​
astra-53 [7]

Answer:

1.5 Amp is rated for 5 W so it would not be possible

7 0
3 years ago
A box of mass 20kg is pulled up an inclined plane by a force of 285N. Given that the value of the incline angle is 30 degrees an
docker41 [41]

Given :

Mass of box , m = 250 kg.

Force applied , F = 285 N.

The value of the incline angle is 30°.

the coefficient of dynamic friction is \mu=0.72 .

To Find :

The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.

Solution :

Net force applied in box is :

F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N

Acceleration , a=\dfrac{F}{m}=\dfrac{60.29}{20}=3.01\ m/s^2.

By equation of motion :

v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s

Therefore, the speed of box is 12.04 m/s.

Hence, this is the required solution.  

6 0
3 years ago
Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g
tekilochka [14]

Answer: a) It will take more time to return to the point from which it was released

Explanation: To determine how long it takes for the ball to return to the point of release and considering it is a free fall system, we can use the given formula:

d=v_{0}.t + \frac{1}{2} .a.t^{2}, where:

d is the distance the ball go through;

v₀ is the initial velocity, which is this case is 0 because he releases the ball;

a is acceleration due to gravity;

t is the time necessary for the fall;

Suppose <em>h</em> is the height from where the ball was dropped.

On Earth:

h=0.t + \frac{1}{2}.10.t^{2}

h = 5t²

t_{T} = \sqrt{\frac{h}{5} }

On the other planet:

h =  0.t + \frac{1}{2}.30.t^{2}

h = 15.t²

t_{P} = \sqrt{\frac{h}{15} }

Comparing the 2 planets:

\frac{t_{T} }{t_{P} } = \frac{\sqrt{\frac{h}{5} } }\sqrt{{\frac{h}{15} } }

\frac{t_{T} }{t_{P} } = \sqrt{3}  or t_{T} = \sqrt{3}.t_{P}

Comparing the two planets, on the massive planet, it will take more time to fall the height than on Earth. In consequence, it will take more time to return to the initial point, when it was released.

5 0
3 years ago
A steeper incline plane will require _____ Force​
ExtremeBDS [4]

Answer:

wind

Explanation:

8 0
3 years ago
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