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2 years ago

If the mass of the object is doubled and the speed is halved then kinetic energy will change by a factor of:

Physics

1 answer:

madam [21]2 years ago

3 0

Answer:

kinetic energy will change by a factor of 1/2

Option C) 1/2 is the correct answer

Explanation:

Given the data in the question;

we know that;

Kinetic energy = 1/2.mv²

given that mass of the object is doubled; m1 = 2m

speed is halved; v1 = V/2

Now, New kinetic energy will be; 1/2.m1v1²

we substitute

Kinetic Energy = 1/2 × 2m × (v/2)²

Kinetic Energy = 1/2 × 2m × (v²/4)

Kinetic Energy = 1/2 × m × (v²/2)

Kinetic Energy = 1/2 [ 1/2mv² ]

Kinetic Energy = 1/2 [ KE ]

Therefore; kinetic energy will change by a factor of 1/2

Option C) 1/2 is the correct answer

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A wire, of length L = 4.1 mm, on a circuit board carries a current of I = 1.96 μA in the j direction. A nearby circuit element g

tamaranim1 [39]

Answer:

The magnitude of the magnetic field B is 5.921 T.

Explanation:

Given that,

Length = 4.1 mm

$B_{x}=4.9\ G$

$B_{y}=2.3\ G$

$B_{z}=2.4\ G$

Current $I = 1.96\ mu A$

We need to calculate the magnetic field

Using formula of magnetic field

$B=\sqrt{B_{x}^2+B_{y}^2+B_{z}^2}$

Put the value into the formula

$B=\sqrt{(4.9)^2+(2.3)^2+(2.4)^2}$

$B=5.921\ T$

Hence, The magnitude of the magnetic field B is 5.921 T.

6 0

3 years ago

Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of

LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

$U=k\frac{q_1q_2}{r_{1,2}}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

$U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}$ (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

$U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J$

The electric potential energy between the three charges is 80.91 J

7 0

3 years ago

A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o

Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let $q_1$ and $q_2$ be the charges such that

$q_1+q_2=4.7$

and Force between charge particles is given by

$F=\frac{kq_1q_2}{r^2}$

$4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}$

$q_1\cdot q_2=0.522$

put the value of $q_1$

$q_2\left ( 4.7-q_2\right )=0.522$

$q_2^2-4.7q_2+0.522=0$

$q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}$

$q_2=0.114 C$

thus $q_1=4.586 C$

3 0

3 years ago

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Citrus2011 [14]

Definitely not the last 2. My bet is on the first option. If it is wrong don't hit me please...

8 0

3 years ago

200 pages about string thirory

joja [24]

Answer:

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Explanation:

7 0

3 years ago

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