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3 years ago

A 5.0 kg object moving at 5.0 m/s. KE = mv2 times 1/2

Physics

1 answer:

steposvetlana [31]3 years ago

5 0

Answer: KE = 62.5J

Explanation:

Given that

Mass of object = 5kg

kinetic energy KE = ?

velocity of object = 5m/s

Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.

i.e K.E = 1/2mv^2

KE = 1/2 x 5kg x (5m/s)^2

KE = 0.5 x 5 x 25

KE = 62.5J

Thus, the object has 62.5 joules of kinetic energy.

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4. The bar has cross-sectional area A and modulus of elasticity E. If an axial force F directed toward the right is applied at C

aniked [119]

Answer:

a) ΔL/L = F / (E A), b) $L_{f}$ = L (1 + L F /(EA) )

Explanation:

Let's write the formula for Young's module

E = P / (ΔL / L)

Let's rewrite the formula, to have the pressure alone

P = E ΔL / L

The pressure is defined as

P = F / A

Let's replace

F / A = E ΔL / L

F = E A ΔL / L

ΔL / L = F / (E A)

b) To calculate the elongation we must have the variation of the length, so the length of the bar must be a fact. Let's clear

ΔL = L [F / EA]

$L_{f}$ -L = L (F / EA)

$L_{f}$ = L + L (F / EA)

$L_{f}$ = L (1 + L (F / EA))

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3 years ago

Prisms seperate __ light, such as that from the Sun by wavelength

natka813 [3]

Answer: white

Explanation:

prisms separate white light, such as that from the sun by wavelength.

Brainliest

3 0

3 years ago

Colonel John P. Stapp, USAF, participated in studying whether a jet pilot could survive emergency ejection. On March 19, 1954, h

PolarNik [594]

Answer:

(a) a = - 201.8 m/s²

(b) s = 197.77 m

Explanation:

(a)

The acceleration can be found by using 1st equation of motion:

Vf = Vi + at

a = (Vf - Vi)/t

where,

a = acceleration = ?

Vf = Final Velocity = 0 m/s (Since it is finally brought to rest)

Vi = Initial Velocity = (632 mi/h)(1609.34 m/ 1 mi)(1 h/ 3600 s) = 282.53 m/s

t = time = 1.4 s

Therefore,

a = (0 m/s - 282.53 m/s)/1.4 s

a = - 201.8 m/s²

(b)

For the distance traveled, we can use 2nd equation of motion:

s = Vi t + (0.5)at²

where,

s = distance traveled = ?

Therefore,

s = (282.53 m/s)(1.4 s) + (0.5)(- 201.8 m/s²)(1.4 s)²

s = 395.54 m - 197.77 m

s = 197.77 m

6 0

3 years ago

Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m

AleksandrR [38]

Answer:

Part a)

$10\hat i + 15\hat j = \vec v$

Part b)

$\Delta K = 500 J$

Explanation:

As we know that there is no external force on the system of two masses so here total momentum of the system will remains conserved

so we can say

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$(2kg)(15\hat i + 30 \hat j) + (2 kg)(-10\hat i + 5\hat j) = 2kg(-5\hat i + 20\hat j) + 2\vec v$

$5\hat i + 35\hat j = (-5\hat i + 20\hat j) +\vec v$

$10\hat i + 15\hat j = \vec v$

Part b)

magnitude of the initial speed of A = $\sqrt{15^2 + 30^2} = 33.54 m/s$

magnitude of the initial speed of B = $\sqrt{10^2 + 5^2} = 11.18 m/s$

magnitude of final speed of A = $\sqrt{5^2 + 20^2} = 20.61 m/s$

magnitude of final speed of B = $\sqrt{10^2 + 15^2} = 18.03 m/s$

Now change in total kinetic energy is given as

$\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)$

$\Delta K = (\frac{1}{2}2(33.54)^2 + \frac{1}{2}2(11.18)^2) - (\frac{1}{2}2(20.61)^2 + \frac{1}{2}2(18.03)^2)$

$\Delta K = 500 J$

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3 years ago

The diagram shows what happens to a system undergoing an adiabatic process.

posledela

The answer is:
B. X: Work is done to the system and temperature increases.
Y: Work is done by the system and temperature decreases.

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4 years ago

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