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Reptile [31]
3 years ago
15

If a dog ran at 5 m/s how far would it run in 45 s

Physics
2 answers:
saveliy_v [14]3 years ago
8 0

Answer:

225 meters

Explanation:

If it is running 5 meters per seconds, and it ran 45 seconds, you would need to multiple the seconds it ran by the speed per second: 5 • 45

5 • 45 = 225

Virty [35]3 years ago
3 0

Answer:

225 m//s

Explanation:

It ran 5m/s so 5x45=225

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sergij07 [2.7K]

Answer: Yes Because it matches with the mass and the amount of force Hope this helps :>

Explanation:

3 0
3 years ago
The wheel on an upside-down bicycle moves through 17.3 rad in 5.41 s. What is the wheel’s angular acceleration if its initial an
user100 [1]

Answer:

0.406 rad/s2

Explanation:

We can use the following equation of motion to calculate the angular acceleration α in term of initial angular speed ω = 2.1 rad/s, time t = 5.41 and angles covered θ = 17.3 rad

\theta = \omega_0 t + \alpha t^2/2

17.3 = 2.1*5.41 + \alpha * 5.41^2/2

17.3 = 11.361 + 14.63 \alpha

\alpha = \frac{17.3 - 11.361}{14.63} = 0.406 rad/s^2

7 0
4 years ago
Read 2 more answers
What is the potential difference VB – VA when the I= 1.5 A in the circuit segment below?​
son4ous [18]

Answer:

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Explanation:

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6 0
3 years ago
A flywheel of mass 182 kg has an effective radius of 0.62 m (assume the mass is concentrated along a circumference located at th
musickatia [10]

Answer:

A)5524J,

B) 29.2Nm

Explanation:

This question can be treated using work- energy theorem

Work= change in Kinectic energy

W= Δ KE

Work= difference between the final Kinectic energy and intial Kinectic energy.

We know that

Kinectic energy= 1/2 mv^2 .............eqn(1)

This can be written in term of angular velocity, as

KE= 1/2 I

4 0
3 years ago
Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter
alukav5142 [94]

Answer: 0.258

Explanation:

The resistance R of a wire is calculated by the following formula:

R=\rho\frac{l}{s}    (1)

Where:

\rho is the resistivity of the material the wire is made of. For aluminium is \rho_{Al}=2.65(10)^{-8}m\Omega  and for copper is \rho_{Cu}=1.68(10)^{-8}m\Omega

l is the length of the wire, which in the case of aluminium is l_{Al}=12m, and in the case of copper is l_{Cu}=30m

s is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

s=\pi{(\frac{d}{2})}^{2}  (2) Where d  is the diameter of the circumference.

For aluminium wire the diameter is  d_{Al}=2.5mm=0.0025m  and for copper is d_{Cu}=1.6mm=0.0016m

So, in this problem we have two transversal areas:

<u>For aluminium:</u>

s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}

s_{Al}=0.000004908m^{2}   (3)

<u>For copper:</u>

s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}

s_{Cu}=0.00000201m^{2}    (4)

Now we have to calculate the resistance for each wire:

<u>Aluminium wire:</u>

R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}     (5)

R_{Al}=0.0647\Omega     (6)  Resistance of aluminium wire

<u>Copper wire:</u>

R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}     (6)

R_{Cu}=0.250\Omega     (7)  Resistance of copper wire

At this point we are able to calculate the  ratio of the resistance of both wires:

Ratio=\frac{R_{Al}}{R_{Cu}}   (8)

\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}   (9)

Finally:

\frac{R_{Al}}{R_{Cu}}=0.258  This is the ratio

3 0
3 years ago
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