3 years ago

Which statements about cells are true? (Select all that apply)

Physics

1 answer:

Gemiola [76]3 years ago

7 0

Answer:

c human have eukaryotic cell

You might be interested in

Rachel hits a golf ball into the air. What type of motion is the ball's path?

sladkih [1.3K]

Answer:

omg

the ground is BREAKING

Explanation:

5 0

3 years ago

What physical property makes metal pots good for cooking

bonufazy [111]

The metal conducts the heat, which makes cooking easier.

5 0

3 years ago

The muzzle velocity of a rifle bullet is 709 m s−1along the direction of motion. If the bullet weighs 35 g, and the uncertainty

nydimaria [60]

Answer:

Uncertainty in position of the bullet is $\Delta x=1.07\times 10^{-33}\ m$

Explanation:

It is given that,

Mass of the bullet, m = 35 g = 0.035 kg

Velocity of bullet, v = 709 m/s

The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :

$p=mv$

$p=0.035\times 709=24.81\ kg-m/s$

Uncertainty in momentum is,

$\Delta p=0.2\%\ of\ 24.81$

$\Delta p=0.049$

We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :

$\Delta p.\Delta x\geq \dfrac{h}{4\pi}$

$\Delta x=\dfrac{h}{4\pi \Delta p}$

$\Delta x=\dfrac{6.62\times 10^{-34}}{4\pi \times 0.049}$

$\Delta x=1.07\times 10^{-33}\ m$

Hence, this is the required solution.

7 0

3 years ago

Suppose a ball is thrown vertically upwards from a position P above the ground. It rises to the highest point Q and returns to t

NikAS [45]

Answer:

Net displacement = 0

Distance traveled = 2PQ <_up and down

Explanation:

5 0

3 years ago

Find the unknown mass of the block 1 needed to balance the bar. Assume that the mass of the bar is negligible. Block 1 is locate

Natasha2012 [34]

Answer:

Incomplete question: The masses of the blocks m₂ = 1.5 kg and m₃ = 2 kg

Explanation:

Given data:

L₁ = length = 0.85 m

L₂ = 0.25 m

L₃ = 0.5 m

m₂ = 1.5 kg

m₃ = 2 kg

Question: Find the unknown mass of the block 1 needed to balance the bar, m₁ = ?

The torque is zero (intermediate point of the bar)

$-m_{1} gL_{1} +m_{2} gL_{2} +m_{3} gL_{3} =0$

Is negative because mass 1 is to the left of the coordinate system (see the diagram)

$m_{1} =\frac{m_{2} gL_{2}+m_{3} gL_{3} }{L_{1} } =\frac{(1.5*0.25)+(2*0.5)}{0.85} =1.6176kg$

4 0

3 years ago