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3 years ago

Determine the mechanical energy of this object: 1-kg ball rolls on the ground at 2 m/s

1 answer:

5 0

The potential energy would be zero. Only kinetic energy is present in this case. To find out what the answer is we do the equation: mv^2/2 soo...

KE =mv^2/2
KE= 1(2^2)/2 which the answer will come up by 2 Joules.

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An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o

Alekssandra [29.7K]

Answer:

$0.53m/s^2$

Explanation:

We are given that

Mass of wire=m=3.6 g= $3.6\times 10^{-3} kg$

1 kg=1000 g

Length of wire=l=1.6 m

Mass of object=m'=3 kg

Time,t=60.1 ms= $60.1\times 10^{-3} s$

$1 ms=10^{-3} s$

Speed,v= $\frac{distance}{time}=\frac{1.6}{60.1\times 10^{-3}}=26.62 m/s$

$g=\frac{v^2m}{m'l}$

Using the formula

$g=\frac{(26.62)^2\times 3.6\times 10^{-3}}{3\times 1.6}=0.53m/s^2$

8 0

3 years ago

A charge +1.9 μC is placed at the center of the hollow spherical conductor with the inner radius 3.8 cm and outer radius 5.6 cm.

Archy [21]

To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.

PART A)

The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:

$Q_{int}=-Q1=-1.9*10^{-6} C$ . This is the total charge on the inner surface of the conducting shell.

PART B)

The positive charge (of the same value) on the external surface of the conducting shell is:

$Q_{ext}=+Q_1=1.9*10^{-6} C$

The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,

$Q_{ext, Total}=Q_2+Q_{ext}$

$Q_{ext, Total}=1.9+3.8$

$Q_{ext, Total}=5.7 \mu C$

5 0

3 years ago

How much time does it take a person to walk 12 km north at a velocity of 6.5 km/hr

Stella [2.4K]

1.8461 km/hr Well i need more characters so i might as well type a beautiful sentence for you to read and waste your time on.

3 0

3 years ago

How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would

professor190 [17]

Answer: 1037 miles per hour

Explanation: In order to see the sun in the same position in the sky, you would have to travel against the speed of rotation of the earth, because this is what causes the sun to appear in a constantly changing position.

Because of this, we will have to calculate the speed of rotation of the earth. To get started, we must know the circumference of the earth. Assuming the circumference formula for a sphere,

$Circumference=2\pi R$

Where R is the radius of the earth, we find that the perimeter of the earth is approximately 24881 miles. The equation to calculate speed is given by

$v=\frac{Distance}{Time}$

Because the earth completes one rotation in 24 hours, we have to find the speed of rotation as the perimeter of the earth divided by 24 hours.

The obtained result is 1037 miles per hour.

You would have to travel at 1037 miles per hour in the direction opposite to the direction the rotation is ocurring in.

5 0

3 years ago

Currents during lightning strikes can be up to 50000 A (or more!). We can model such a strike as a 47500 A vertical current perp

Rufina [12.5K]

Answer:

57 N

Explanation:

Force on a current carrying conductor in a magnetic field

B = 12 X 10⁻⁴ T

= Bil where B is magnetic field , i is current and l is length of conductor

force required = 12 x10⁻⁴ x 47500 x 1

= 57 N

6 0

3 years ago