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2 years ago

A car increases its speed from 4.20 m/s to 8.60 m/s over 3.20 seconds. What is the car’s acceleration?

Physics

1 answer:

katrin2010 [14]2 years ago

5 0

Answer:1.375metre per second square

Explanation: acceleration=(final velocity-initial velocity)÷time

acceleration=(8.6-4.2)÷3.2

Acceleration=4.4÷3.2

Acceleration=1.375 metre per second square

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The magnitudes of two forces are measured to be 120 ± 5 N and 60 ± 3 N. Find the sum and diff erence of the two magnitudes, givi

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62N

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3 years ago

Which statement best define energy

Ksju [112]

Answer:

"The capacity of a system to perform work of any type."

Explanation:

The best statement to describe Energy is:

"The capacity of a system to perform work of any type."

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3 years ago

What happens to the frequency of a wave of you increase the speed of the wave ?

exis [7]

Nothing happens. The frequency is determined at the source,
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3 years ago

How much total energy is dissipated in 10. seconds

noname [10]

Answer : Total energy dissipated is 10 J

Explanation :

It is given that,

Time. t = 10 s

Resistance of the resistors, R = 4-ohm

Current, I = 0.5 A

Power used is given by :

$P=\dfrac{E}{t}$

Where

E is the energy dissipated.

So, E = P t.............(1)

Since, $P=I^2R$

So equation (1) becomes :

$E=I^2Rt$

$E=(0.5\ A)^2\times 4\Omega \times 10\ s$

$E=10\ J$

So, the correct option is (3)

Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

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3 years ago