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Rufina [12.5K]
2 years ago
7

At what speed does a 1,248 kg compact car have the same kinetic energy as a 18,777 kg truck going 26 km/h?

Physics
1 answer:
notka56 [123]2 years ago
3 0

The speed of car is 100.8km/h

KE =  \frac{1}{2} m(v \: truck ) {}^{2}

= 0.5 \times 18777 \times  \frac{26}{3.6} \times  \frac{26}{3.6}

= 489708.79j

=  \frac{1}{2}  \times 1248 \times( v \: car) {}^{2}

= 624v {}^{2}

so \: v {}^{2}  =  \frac{489708.7}{624}

= 784

v =  \sqrt{784}

v = 28m/s

v car= 28×3.6

=100.8km/h

Hence, the speed of the car is 100.8km/h

learn more about speed from here:

brainly.com/question/28326855

#SPJ4

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A bowling ball with a mass of 10 kg is moving at a speed of 3 m/s. What is its
jasenka [17]
The answer would be C.30J
3 0
4 years ago
A gold wire has a cross-sectional area of 1.0 cm^2 and a resistivity of 2.8 × 10^-8 Ω ∙ m at 20°C. How long would it have to be
Karo-lina-s [1.5K]

Answer:

Length, l = 3.57 meters

Explanation:

It is given that,

Area of cross-section of gold wire, A=1\ cm^2=0.0001\ m^2

Resistivity of gold wire, \rho=2.8\times 10^{-8}\ \Omega-m

Resistance, R = 0.001 ohms

Resistance in terms of length and area is given by :

R=\rho \dfrac{l}{A}

l=\dfrac{RA}{\rho}

l=\dfrac{0.001\times 0.0001}{2.8\times 10^{-8}}

l = 3.57 meters

So, the length of the wire is 3.57 meters. Hence, this is the required solution.

4 0
3 years ago
When a charge of 10 C flows past any point along a circuit in 5 seconds, the current at that point would be _______ A.
aleksandrvk [35]

Answer:

2 amps

Explanation:

I=Q/t

I=10/5

I=2 Amps

4 0
4 years ago
A simple generator has a square armature 9.0 cm on a side. The armature has 95 turns of 0.59-mm-diameter copper wire and rotates
Marrrta [24]

Answer:

f = 3.102 Hz

Explanation:

In this case you have that the required voltage is the maximum induced emf produced by the rotating generator.

In order to calculate the frequency of rotation of the generator that allows one to obtain 12.0V you use the following formula:

emf_{max}=NBA\omega    (1)

N: turns of the armature = 95

B: magnitude of the magnetic field = 0.800T

A: area of the square armature = (9.0cm)^2 = (0.09m)^2 = 8.1*10^-3 m^2

emf_max = 12.0V

w: angular frequency

you solve the equation (1) for w:

\omega=\frac{emf_{max}}{NBA}=\frac{12.0V}{(95)(0.800T)(8.1*10^{-3}m^2)}\\\\\omega=19.49\frac{rad}{s}

Then, the frequency is:

f=\frac{\omega}{2\pi}=\frac{19.49rad/s}{2\pi}=3.102Hz

7 0
3 years ago
A single loop of current is immersed in an externally applied uniform magnetic field of 3 Tesla oriented in the positive y direc
vlabodo [156]

Answer:

mu=12Tm^2

Explanation:

the magnetic moment mu of a single loop is given by:

\mu = I A B

where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:

\mu=(0.5A)(4m*2m)(3T)=12Tm^2

hope this helps!!

6 0
3 years ago
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