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2 years ago

7

If there were no dark energy in the universe, the value of __________ would solely determine the evolution and fate of the unive

rse.

1 answer:

Talja [164]2 years ago

5 0

Answer:

Ωmass

Explanation:

If there were no dark energy in the universe, the value of Ωmass would determine the evolution and fate of the universe.

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How would the force of a test charge change if the electric field is doubled?

Stolb23 [73]

<span>The force would double.</span>

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3 years ago

Read 2 more answers

A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule

saw5 [17]

Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

$m_1v_1+m_2v_2=0$

$170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0$

$v_2=-0.17\ m/s$

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

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3 years ago

A model rocket is launched with an initial upward velocity of.

REY [17]

Answer:
A model rocket is launched with an initial upward velocity of 215 ft/s.

Explanation:

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2 years ago

Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having

igomit [66]

Answer:

(a) A = m/s^3, B = m/s.

(b) dx/dt = m/s.

Explanation:

(a)

$x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s$

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.

(b) $\frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s$

This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.

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3 years ago

A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what

irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

$\Delta L=\frac{PL}{AE}$

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

$A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2$

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

$0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg$

Mass of the climber = 69.38 kg

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3 years ago