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Flura [38]
2 years ago
14

Ex 10: My dog runs at 6 m/s for 18 meters. How long did she run for?

Physics
1 answer:
Hunter-Best [27]2 years ago
7 0
She ran for 3s

Put 18/6 because in order to find how long she ran for you need to divide the distance by the meters ran, once you do that you will get 3.
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What is the purpose of a free body diagram
Amiraneli [1.4K]
<span>A free body diagram is a representation of how the forces that are acting on a point or particle interact. You place your point at the origin and then draw your forces.


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4 0
3 years ago
Read 2 more answers
How far will a car travel in 25 seconds at 10m/ min
Kryger [21]
Time = 25s
speed = 10m/min
= 10m / 60
= (1/6)m/s


distance = speed × time
= 25 × (1/6)
=4.167m
7 0
3 years ago
If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
3 years ago
Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m
denis23 [38]

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

8 0
2 years ago
Which of the following is a chemical change?
klemol [59]

a). Water is still H₂O after it freezes.

b). Ice is still H₂O after it melts.

c). Wire is still Cu when it's bent.

d). Paper combines with the O₂ in the air, and turns into
     a lot of new compounds when it burns.

3 0
3 years ago
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