How many miles is the moon from the earth

Which has more kinetic energy, a 4.0 kg bowling ball moving at 1.0 m/s or a 1.0 kg

jolli1 [7]

Answer:

The kinetic energy of bocce ball is more.

Explanation:

Given that,

Mass of a bowling ball, m₁ = 4 kg

Speed of the bowling ball, v₁ = 1 m/s

Mass of bocce ball, m₂ = 1 kg

Speed of bocce ball, v₂ = 4 m/s

We need to say which has more kinetic energy.

The kinetic energy of an object is given by :

$E=\dfrac{1}{2}mv^2$

Kinetic energy of the bowling ball,

$E_1=\dfrac{1}{2}m_1v_1^2\\\\E_1=\dfrac{1}{2}\times 4\times (1)^2\\\\E_1=2\ J$

The kinetic energy of the bocce ball,

$E_2=\dfrac{1}{2}m_2v_2^2\\\\E_2=\dfrac{1}{2}\times 1\times (4)^2\\\\E_2=8\ J$

So, the kinetic energy of bocce ball is more than that of bowling ball.

5 0

2 years ago

What is the total resistance in the circuit? (include unit in answer - ohms) I WILL GIVE BRAINLIEST!!

VashaNatasha [74]

YES I TOOK THIS ! The answer is 25.79.

8 0

2 years ago

Read 2 more answers

A 1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s.

iren [92.7K]

Explanation:

Given that,

Mass of the ball, m = 1.2 kg

Initial speed of the ball, u = 10 m/s

Height of the floor from ground, h = 32 m

(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 32}$

v = -25.04 m/s (negative as it rebounds)

The impulse acting on the ball is equal to the change in momentum. It can be calculated as :

$J=m(v-u)$

$J=1.2\times (-25.04-10)$

J = -42.048 kg-m/s

(b) Time of contact, t = 0.02 s

Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :

$J=\dfrac{F}{t}$

$F=J\times t$

$F=42.048\times 0.02$

F = 0.8409 N

Hence, this is the required solution.

5 0

2 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

2 years ago

A coil of area 0.320 m2 is rotating at 100 rev/s with the axis of rotation perpendicular to a 0.430 T magnetic field. If the coi

weqwewe [10]

Answer:

The maximum emf generated in the coil is 60527.49 V

Explanation:

Given;

area of coil, A = 0.320 m²

angular frequency, f = 100 rev/s

magnetic field, B = 0.43 T

number of turns, N = 700 turns

The maximum emf generated in the coil is calculated as,

E = NBAω

where;

ω is the angular speed = 2πf

E = NBA(2πf)

Substitute in the given values and solve for E

E = 700 x 0.43 x 0.32 x 2π x 100

E = 60527.49 V

Therefore, the maximum emf generated in the coil is 60527.49 V

8 0

3 years ago