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3 years ago

Please help. I'll give brainliest :)

Physics

2 answers:

makkiz [27]3 years ago

6 0

Answer:

I think It should be A ........

rusak2 [61]3 years ago

4 0

Answer:

I think the answer might be A or B

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A stone dropped from the top of a building reaches the ground with a velocity of 49ms¹. If the acceleration due to gravity is 9.

bezimeni [28]

Explanation:

Given:

v₀ = 0 m/s

v = 49 m/s

a = 9.8 m/s²

Find: t

v = at + v₀

49 m/s = (9.8 m/s²) t + 0 m/s

t = 5 s

6 0

3 years ago

Which isotope has 16 neutrons and is used to study nucleotides and nucleic acids?

qaws [65]

When naming isotopes, the number beside it indicates the mass number. Now mass number is the sum of protons and neutrons.

We know that the mass number of a stable Sulfur atom is 32, so we can rule out C as an isotope. The atomic number of Sulfur is 16 and that means that there are 16 protons. Now if we subtract 16 from 33, then that means there are 17 neutrons. So we can rule out D.

Phosphorus on the other hand has an atomic number of 15. If we subtract 15 from 31 we will have 16. That means that Phosphorus-31 has 16 neutrons.

The answer would then be A. Phosphorus 31.

7 0

4 years ago

When the fission of uranium-235 is carried out, about 0.1 percent of the mass of the reactants is lost during the reaction.

Dmitry [639]

The lost mass is converted into energy.

7 0

3 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

8 0

3 years ago

Suppose a star the size of our Sun, but with mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 7.0 days.

kvasek [131]

Answer:

Its rotation will be 3.89x10⁴ rad/s.

Explanation:

We can find the rotation speed by conservation of the angular momentum:

$L_{i} = L_{f}$

$I_{i}\omega_{i} = I_{f}\omega_{f}$ (1)

The initial angular speed is:

$\omega_{i} = \frac{1 rev}{7 d} = 0.14 \frac{rev}{d}$

The moment of inertia (I) of a sphere is:

$I = \frac{2}{5}mr^{2}$ (2)

Where m is 9 times the sun's mass and r is the sun's radius

By entering equation (2) into (1) we have:

$\frac{2}{5}m_{i}r_{i}^{2}\omega_{i} = \frac{2}{5}m_{f}r_{f}^{2}\omega_{f}$

$9m_{sun}(696342 km)^{2}0.14\frac{rev}{d} = \frac{3}{4}9m_{sun}(13 km)^{2}\omega_{f}$

$\omega_{f} = \frac{4}{3}*0.14 \frac{rev}{d}(\frac{696342 km}{13 km})^{2} = 5.36 \cdot 10^{8} \frac{rev}{d}*\frac{1 d}{24 h}*\frac{1 h}{3600 s}*\frac{2\pi rad}{1 rev} = 3.89 \cdot 10^{4} rad/s$

Hence, its rotation will be 3.89x10⁴ rad/s.

I hope it helps you!

3 0

4 years ago