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garri49 [273]
3 years ago
7

At 9:00 on Saturday morning, two bicyclists heading in opposite directions pass each other on a bicycle path.The bicyclist headi

ng north is riding 5 km/hour faster than the bicyclist heading south. At 10:45, they are 47.25 km apart. Find the two bicyclists’ rates (speed/velocity).
Physics
2 answers:
guajiro [1.7K]3 years ago
5 0

Answer:

northbound bicyclist = 16 km/h; southbound bicyclist = 11 km/h

<em>Hope this helps</em>

<em>-Amelia The Unknown</em>

notka56 [123]3 years ago
3 0
Between 9:00 am and 10:45 am, there have been 1 hour and 45 minutes or 1.75 hours have passed. Let x be the speed of the slower cyclist and x+ 5 be the rate of the second cyclist. The given situation is best represented through the equation below,
                                 x(1.75) + (x + 5)(1.75) = 47.25 km
The value of x from the equation is 11. Thus, the two bicyclists' rates are 11 km/h and 16 km/h. 
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A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer to the nearest tenth.The
Galina-37 [17]
The speed of the runner is 300 m /38 seconds. You can simplify this answer to be about 7.9 m/s
7 0
3 years ago
1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

where

mg is the weight of the object

\Delta h is the change in height

For the box in this problem,

mg = 185 N

\Delta h = 0.800 m

Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

W=(825)(35)=28875 J

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

W' = 2(28875)=57750 J

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

where

m = 0.180 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

W=(1.76)(2.5)=4.4 J

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

Solving for m, we find

m=\frac{W}{g \Delta h}

And substituting the numerical values, we find the mass of the box:

m=\frac{7000}{(9.8)(1.2)}=595.2 kg

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

W=mg \Delta h

Where \Delta h is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

5 0
3 years ago
We know the frequency range of certain sounds are: 400-560 Hz, what are the ranges of wavelength in meters when the signal trans
Ksivusya [100]

Answer:

Range of wavelength will be 5.35\times 10^5m to 7.5\times 10^5m

Explanation:

We have given range of frequency is 400-560 Hz

Speed of the light c=3\times 10^8m/sec

We have to find the range of the wavelength of signal transmitted

Ween know that velocity is given by v=\lambda f, here \lambda is wavelength and f is frequency

So for 400 Hz frequency wavelength will be \lambda =\frac{c}{f}=\frac{3\times 10^8}{400}=7.5\times 10^5m

And wavelength for frequency 560 Hz \lambda =\frac{c}{f}=\frac{3\times 10^8}{560}=5.35\times 10^5m

So range of wavelength will be 5.35\times 10^5m to 7.5\times 10^5m

8 0
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A parked car begins to roll down a hill, what can you conclude from that observation?
Fynjy0 [20]

Answer:

its The rolling friction is greater than the force of the car’s weight against the hill.

and A force was required to start the car rolling.

Explanation:

3 0
3 years ago
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.503 X 104 T) at a distance of 15
dem82 [27]

Answer:

37.725 A

Explanation:

B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T

r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m

i = magnitude of current flowing through the wire = ?

Magnetic field by a long wire is given as

B = \frac{\mu _{o}}{4\pi }\frac{2i}{r}

Inserting the values

0.503\times 10^{-4} = (10^{-7})\frac{2i}{0.15}

i = 37.725 A

6 0
3 years ago
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