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lora16 [44]
3 years ago
11

A dog walks a distance of 55.5 meters in 120 seconds. What was its speed?

Physics
1 answer:
Nitella [24]3 years ago
8 0

Answer: ...

Explanation: 1: The dog is walking 0.4625MPS (Meters per second. M/S)

To calculate speed you divide the distance by time so 55.5 Meters/120 Seconds is 0.4625 MPS

2: The car traveled 15 KM

To calculate speed you need to multiply the speed by time. (Since it's multiplication it can be put in any order) So 600 Secondsx25 Meters is 15,000 M which is 15 KM.

3: It would take 8 Minutes and 20 Seconds to travel 1500 Meters.

To calculate the time you multiply Speed by distance (Order doesn't count.)

So 1500 Metersx3 MPS is 8 Minutes and 20 seconds.

4: It was going 2.7775 MPS.

To calculate the speed you divide the distance over time. So 9999 Meters/3600 seconds is 2.7775 MPS.

5: It was going 1.1222... (Infinite)

To calculate the speed you divide the distance over time

So 8080 Meters/ 7200 Seconds is 1.222...

6: It would cover 1.375 Meters.

To calculate the Distance you multiply Speed by time (Order doesn't count.)

So 10.25 Secondsx5.5 MPS is 1.375 Meters.

7: It would take 1 Minute and 31 Seconds to travel 10 Meters.

To calculate the time you multiply Speed by distance (Order doesn't count.)

So 10 Metersx0.11 MPS is 1 Minute and 31 seconds

Hopes this helps because this took over 10 minutes.

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4. A car accelerating at 60 meters/second is hit in an accident by a bus. The net force exerted on
Licemer1 [7]

Answer:

500kg

Explanation:

mass = newtons/force divided by the acceleration rate

m = 30,000/60

m = 500

3 0
2 years ago
Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t
earnstyle [38]
1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
B= \frac{\mu_0I}{2 \pi r}
where
\mu_0 is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance 
r=40.0 cm=0.40 m, 
which is the distance at which the other wire is located:
B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N

2) direction of the force: 
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
6 0
2 years ago
A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.
Crank

Answer: 1000 Hz

Explanation:

You can calculate frequency by dividing velocity by wavelength

Frequency = velocity/wavelength

Find velocity first.

900 m/3 s = 300 m/s

Plug values in to find frequency.

F = (300 m/s)/0.3 m

F = 1000 Hz

8 0
2 years ago
A star is moving away from an observer at 1% of the speed of light. At what wavelength would the observer find an emission line
Ivan

Answer:

  λ = 5940 Angstroms

Explanation:

This is an exercise of the relativistic Doppler effect

        f’= f  √((1- v / c) / (1 + v / c))

Where the speed in between the strr and the observer is positive if they move away

Let's use the relationship

         c = λ f

         f = c /λ

We replace

              c /λ’ = c /λ  √ ((1- v / c) / (1 + v / c))

              λ = λ’ √ ((1- v / c) / (1 + v / c))

Let's calculate

             v = 0.01 c

             v = 0.01 3 10⁸

             v=  3 10⁶ m / s

             λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]

             λ = 6000 √ [0.99 / 1.01]

             λ = 5940 Angstroms

6 0
3 years ago
If 128 v is required to push 4.00 A of current though a resistor, what is the resistance
prisoha [69]

Answer:

32.0 Ω

Explanation:

Ohm's law:

V = IR

128 V = (4.00 A) R

R = 32.0 Ω

8 0
2 years ago
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