All categories

3 years ago

(06.04 MC) Which of the following best describes a human egg cell? (3 points)

Physics

1 answer:

Vinil7 [7]3 years ago

4 0

Answer: A haploid cell formed in the female uterus

You might be interested in

What is precision?what is precision?

RUDIKE [14]

Answer:

According to Oxford Dictionaries "Precision" means "the quality, condition, or fact of being exact and accurate."

Explanation:

Hope this helps! :)

6 0

2 years ago

Read 2 more answers

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h

hram777 [196]

Answer:

$v_{2}=3.5 m/s$

Explanation:

Using the conservation of energy we have:

$\frac{1}{2}mv^{2}=mgh$

Let's solve it for v:

$v=\sqrt{2gh}$

So the speed at the lowest point is $v=7 m/s$

Now, using the conservation of momentum we have:

$m_{1}v_{1}=m_{2}v_{2}$

$v_{2}=\frac{1*7}{2}$

Therefore the speed of the block after the collision is $v_{2}=3.5 m/s$

I hope it helps you!

8 0

3 years ago

A planet orbits a star, in a year of length 3.37 x 107 s, in a nearly circular orbit of radius 1.04 x 1011 m. With respect to th

PIT_PIT [208]

Answer

Given,

Time period of star,T = 3.37 x 10⁷ s

Radius of circular orbit,R = 1.04 x 10¹¹ m

a) Angular speed of the planet

$\omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}$

$\omega = 1.864\times 10^{-7}\ rad/s$

b) tangential speed

$v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}$

v = 1.94 x 10⁴ m/s

c) centripetal acceleration magnitude

$a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}$

a = 3.62 x 10⁻³ m/s²

8 0

3 years ago

a crowbar of 2 meter is used to lift an object of 800N if the effort arm is 160cm , calculste the effort applied

Vitek1552 [10]

Answer:

200 N

Explanation:

The crowbar is 2 meter, or 200 cm. The effort arm is 160 cm, so the moment arm of the object is 40 cm.

(800 N) (40 cm) = F (160 cm)

F = 200 N

5 0

3 years ago

A toy car having mass m = 1.50 kg collides inelastically with a toy train of mass M = 3.60 kg. Before the collision, the toy tra

baherus [9]

Answer:

$v_{f}$ = 3,126 m / s

Explanation:

In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.

the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s

Before the crash

p₀ = m v₁₀ + M v₂₀

After the inelastic shock

$p_{f}$ = m $v_{1f}$ + M $v_{2f}$

p₀ = $p_{f}$

m v₀ + M v₂₀ = m $v_{1f}$ + M $v_{2f}$

We cleared the end of the train

M $v_{2f}$ = m (v₁₀ - v1f) + M v₂₀

Let's calculate

3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45

$v_{2f}$ = (-3.9 + 8.82) /3.60

$v_{2f}$ = 1.36 m / s

As we can see, this speed is lower than the speed of the car, so the two bodies are joined

set speed must be

m v₁₀ + M v₂₀ = (m + M) $v_{f}$

$v_{f}$ = (m v₁₀ + M v₂₀) / (m + M)

$v_{f}$ = (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)

$v_{f}$ = 3,126 m / s

4 0

4 years ago