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3 years ago

A solid ball is rolling along a horizontal surface at 3.7 m/s when it encounters an upward

Physics

1 answer:

lys-0071 [83]3 years ago

4 0

Answer:

h = 0.697 [m]

Explanation:

To solve this problem we must use the energy conservation theorem, where it tells us that kinetic energy is converted to potential energy or vice versa.

$E_{kinet}=E_{pot}$

where:

Ekinet = kinetic energy [J]

Epot = potential energy [J]

$\frac{1}{2}*m*v^{2}=m*g*h\\0.5*(3.7)^{2} =9.81*h\\h =(6.845)/9.81\\h = 0.697 [m]$

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As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

2 years ago

What are all stars made of

ivolga24 [154]

Stars are huge celestial bodies made mostly of hydrogen and helium that produce light and heat from the churning nuclear forges inside their cores. Aside from our sun, the dots of light we see in the sky are all light-years from Earth. They are the building blocks of galaxies, of which there are billions in the universe. It’s impossible to know how many stars exist, but astronomers estimate that in our Milky Way galaxy alone, there are about 300 billion.

7 0

3 years ago

5. A hollow cylinder of mass m, radius Rc, and moment of inertia I = mRc2 is pushed against a spring (with spring constant k) co

Makovka662 [10]

Explanation:

(a) Draw a free body diagram of the cylinder at the top of the loop. At the minimum speed, the normal force is 0, so the only force is weight pulling down.

Sum of forces in the centripetal direction:

∑F = ma

mg = mv²/RL

v = √(g RL)

(b) Energy is conserved.

EE = KE + RE + PE

½ kd² = ½ mv² + ½ Iω² + mgh

kd² = mv² + Iω² + 2mgh

kd² = mv² + (m RC²) ω² + 2mg (2 RL)

kd² = mv² + m RC²ω² + 4mg RL

kd² = mv² + mv² + 4mg RL

kd² = 2mv² + 4mg RL

kd² = 2m (v² + 2g RL)

d² = 2m (v² + 2g RL) / k

d = √[2m (v² + 2g RL) / k]

8 0

2 years ago

In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = $6.68 * 10^{14}Hz$

6 0

2 years ago

what gas is most abundant greenhouse gas ? a) ozone b) chlorofluorocarbon c) carbon dioxide d) methane e) water vapor

jek_recluse [69]

i believe the answer is water vapor

hope this helps

5 0

3 years ago

Read 2 more answers