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3 years ago

The equation E = MC squared relates energy and 1. Force 2. Gravity 3. Mass 4. Work

Physics

2 answers:

Sladkaya [172]3 years ago

8 0

The equation relates Energy to mass

Ganezh [65]3 years ago

7 0

The equation relates energy to mass

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For a vehicle to negotiate a banked curve in poor weather conditions, where the force of friction f = 0, for a given velocity v

djverab [1.8K]

Answer:

$R=240m$

Explanation:

From the question we are told that:

Velocity $v=25m/s$

Force of friction f = 0

Angle $\theta=15$

Generally the equation for Radius of curvature is mathematically given by

$R=frac{v^2}{tan\theta *g}$

$R=frac{25^2}{tan 15 *9.81}$

$R=240m$

7 0

2 years ago

A truck heading east has an initial velocity of 6/ms. It accelerates at 2/ms2 for 12 seconds. What distance does the truck trave

frez [133]

Answer:

216m

Explanation:

As we know S=ut+1/2at^2

S=6×12+1/2×2×12×12

=72+144

=216 m

4 0

2 years ago

When vibrational motion in an object increases, which is a true statement?

ss7ja [257]

Hello friend!!

We know that kinetic energy is the energy possessed due to the motion of the object. And we know if the object is in a fast motion then the temperature would be high, whereas if the object is slow in motion then it will have lower temperature. So we know that the kinetic energy is indirectly related to temperature.From our knowledge we can conclude that HIGHER THE TEMPERATURE, HIGHER THE KINETIC ENERGY and LOWER THE TEMPERATURE, LOWER THE KINETIC ENERGY.
Hence, the answer to your question here is,a.kinetic energy, temperature, and thermal energy increase.
Hope it helps!!All the best!!

5 0

3 years ago

A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

iris [78.8K]

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

$T = 2 \pi \sqrt{\frac{m}{k} } \\$

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

$T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} = \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k = 17.71N/m$

Hence the force constant of the spring is 17.71N/m

4 0

2 years ago

On a cold day, a heat pump absorbs heat from the outside air at 14°F (−10°C) and transfers it into a home at a temperature of 86

Leto [7]

Answer:

6.575

Explanation:

T1 = 30C = 30 + 273 = 303 K

T2 = - 10 C = - 10 + 273 = 263 K

The coefficient of performance of heat pump

k = T2 / (T1 - T2)

k = 263 / (303 - 263) = 6.575

7 0

3 years ago