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3 years ago

10

A drag racer starts from rest and accelerates at 10 m/s squared for the

1 answer:

nika2105 [10]3 years ago

6 0

Answer:

54 $\frac{m}{s}$

Explanation:

$v=u+at$

where $u=$ initial velocity

$v=0+(10\frac{m}{s^{2} } )$ × $(5.4s)$

$v=54\frac{m}{s}$

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A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force

Tom [10]

Answer:

$VB - VA = - 33.4$

Explanation:

Generally the workdone in moving the proton is mathematically represented as

$W = KE_f - KE_i$

Where $KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy$

So

$KE_i = \frac{1}{2} m v_a^2$

Here $v_a$ is the velocity at A with value 50 m/s

So

$KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2$

$KE_i = 2.09 *10^{-24} \ J$

Also

$KE_f = \frac{1}{2} m v_b^2$

Here $v_a$ is the velocity at A with value $80 km/s = 80000 m/s$

=> $KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2$

=> $KE_f = 5.34 *10^{-18} \ J$

So

$W = 5.34 *10^{-18} - 2.09 *10^{-24}$

$W = 5.34 *10^{-18} m/s$

Now this workdone is also mathematically represented as

$W = q * V$

So

$q * V = 5.34 *10^{-18}$

Here $q = 1.60*10^{-19} C$

So

$V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}$

$V = 33.4 \ V$

Generally proton movement is in the direction of the electric field it means that $VA>VB$

So

$VB - VA = - 33.4$

8 0

3 years ago

How much energy is required to accelerate a golf ball of mass 0.046 kg initially at rest to a speed of 0.75c?

Andrew [12]

The required energy to accelerate the golf ball is $0.232875X10^{16} kg-m/sec^2$ .

<h2>What is Energy?</h2>

Energy, which is observable in the execution of labor as well as in the form of light and heat, is the quantitative quality that is transmitted to an object or to a physical phenomenon in physics.
Energy is a preserved resource; according to the rule of conservation of energy, energy can only be transformed from one form to another and cannot be created or destroyed.
The joule, which is defined as "the energy transmitted to an object by the effort of moving it a length of one metre against a force of one newton," is the standard measure for power in the International System of Units (SI).

Given that,

E= $mc^2$

= $0.046X0.75^2Xc^2$

= $0.232875X10^{16} kg-m/sec^2$ .

Learn more about energy here:

brainly.com/question/13881533

#SPJ4

3 0

2 years ago

Take a close look at the energy transfers and transformations shown in the above diagram. Which type of energy is transformed in

Elanso [62]

Answer:

kinetic energy

Explanation:

a certain amount of energy is transferred by the kick. The ball gains an equal amount of energy, mostly in the form of kinetic energy.

4 0

3 years ago

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The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud

motikmotik

<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = $\frac{Gm_1m_2}{r^2}$

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

and r is the distance between them

Thus r = $\sqrt{\frac{Gm_1m_2}{F} }$

If we substitute the values in the above equation

r = $\sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }$

= 2.46 m

3 0

3 years ago

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HELP ASAP!! 15 POINTS!!

Allisa [31]

Answer:

The bottom one has the highest amplitude

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3 years ago

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