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3 years ago

Find the average speed of a horse that traveled east for 25km in 4 hours.

Physics

1 answer:

MArishka [77]3 years ago

4 0

The direction the horse travels has nothing to do with its speed but...

The horse would be traveling 6.25 km/hr

You simply take the overall travel (25km) and divide it by the total time (4hr)

You get 6.25km as your answer, meaning that every hour the horse traveled 6.25km every hour.

4 hours later the horse has reached 25km

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A railroad car moves under a grain elevator at a constant speed of 3.20 m/s. Grain drops into the car at the rate of 240 kg/min.

dedylja [7]

Answer:

F = 768 N

Explanation:

It is given that,

Speed of the elevator, v = 3.2 m/s

Grain drops into the car at the rate of 240 kg/min, $\dfrac{dm}{dt}=240\ kg/min = 4\ kg/s$

We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :

$F=\dfrac{dp}{dt}$

$F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}$

Since, the speed is constant,

$F=m\dfrac{dv}{dt}$

$F=v\dfrac{dm}{dt}$

$F=3.2\times 240$

F = 768 N

So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.

5 0

3 years ago

Which of the following is a chemical equation that accurately represents what happens when a sulfur and oxygen are produced from

Ksenya-84 [330]

It's c................

5 0

3 years ago

A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i

galben [10]

Answer:

350 F to 100 F it take approx 87.33 min

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling

$\frac{dT}{dt}$ = -k(T-Ta)

$\frac{dy}{dt}$ = $\frac{d}{dt}$ (T(t) -Ta)

= $\frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt}$ = -k(T-Ta)

-ky $\frac{dy}{dt}$ = -ky

T(t) -Ta = (To -Ta) $e^{-kt}$ T(t) = Ta+ (To -Ta) $e^{-kt}$

put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) $e^{-k30}$

k = 0.025575

so here for T(t) = 100F

100 = 70 + ( 350 - 70 ) $e^{-0.025575*t}$

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min

5 0

3 years ago

A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.

Ray Of Light [21]

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>

Mass of the box, m = 3.1 kg
Distance moved by the box, d = 2.5 m
Coefficient of friction, = 0.35
Inclination of the force, θ = 30⁰

Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

$W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)$

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

$W = (0) d \times cos(30)\\\\W = 0 \ J$

The work done by the force of gravity is calculated as follows;

$W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J$

The work done on the box by the normal force is calculated as follows;

$W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J$

Learn more about work done here: brainly.com/question/8119756

8 0

2 years ago

If an exit sign is hanging from the ceiling by three chains each of which has a tension of 15N what is the weight of the sign

kvasek [131]

The weight is 45 N, because the three chains hold the sign, and each contributes 15 N.

Notice that the mass would be the weight/acceleration of gravity, m = 45/9.8 kg. But they ask the weight (force, so Newtons)

3 0

3 years ago