What is the end product of mitosis?
A. Four gametes
B. Two identical nuclei
C. One complete zygote
D. A cell with one set of chromosomes

Answer:

60 km per hour

Explanation:

if you drive for 30 min and go 30km then if you drive for 60 min(1hr) then you would have driven 60km.

Answer:

60

Explanation:

hope this helps

Answer:

Yes

Explanation:

Yes, this is a random error generating because of statistical constraint. We only have finite number of data points. As per this, if we plot our observation we will get a gaussian (inverse bell ) shaped curve with mean equal to central value.

Answer:

Explanation:

The average speed can be found by using the formula

[tex]v = \frac{d}{t} \\ [/tex]

d is the distance

t is the time taken

From the question we have

[tex]v = \frac{100}{10} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

B. A dish of water is placed in the freezer and turns solid

Explanation:

it is a physical change because it can return to its original state by physical means. the other options are chemical changes because things like color change, bubbles or fizzing, and odor change all indicate a chemical change and cannot return to the original state it was in before.

I think it because of UV rays ultra violet ray which can make their colors different

Answer:

Magnitude and direction describe velocity

However if there is only magnitude its mean it is speed.

Fun : probably you have seen speed movie. It could not have director because if speed is directed then it will be velocity lol..

Answer:

a = 0.98J

Explanation:

PE = mgh

PE = 0.1 x 9.8 x 1

PE = 0.98J

(a) The potential energy of the car at the beginning of the experiment is 9.8(m) J.

(b) The kinetic energy of the car for group a is 3.125(m) J.

(c) The kinetic energy of the car for group b is 4.5(m) J.

The potential energy of the car is calculated as follows;

P.E = mgh

where;

P.E = m x 9.8 x 1

P.E = 9.8(m) J

K.E = ¹/₂mv²

K.E = ¹/₂ x m x (2.5)²

K.E = 3.125(m) J

K.E = ¹/₂mv²

K.E = ¹/₂ x m x (3)²

K.E = 4.5(m) J

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Answer:

Water

Explanation:

Answer:

So, water has a specific heat of exactly 1.00 calorie per gram per °C. Because one calorie is equal to 4.18 J, it takes 4.18 J to raise the temperature of one gram of water by 1°C. In joules, water's specific heat is 4.18 J per gram per °C.

The answer is water.

Answer:  002 10.0points Two blocks are arranged at the ends of a massless string as shown in . 96 kg mass has fallen through 0 . 415 m, its downward speed is 1 . 25 m / s.

Explanation: I hoped that helped!!

Answer:

2.5 kg

Explanation:

Answer:

Can you show us the route and then I would be happy to help :)

Explanation:

Answer:

A. 39.2 m/s

B. 78.4 m

Explanation:

Data obtained from the question include:

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

A. Determination of the brick's velocity.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = gt

v = 4 × 9.8

v = 39.2 m/s

Thus, the brick's velocity after 4 s is 39.2 m/s

B. Determination of how far the brick fall in 4 s.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

h = ½gt²

h = ½ × 9.8 × 4²

h = 4.9 × 16

h = 78.4 m

Thus, the brick fall 78.4 m during the time.

The velocity after 4 seconds is 40 m/s and the height is 80 m.

From the equations of motion under gravity;

v = u + gt

V = final velocity of the brick

u = initial velocity of the brick

g = acceleration due to gravity = 10 ms-2

t = time taken

Note that, u = 0 m/s because the brick was dropped from a height.

v = 0 + 10(4)

v = 40 m/s

The height is obtained from;

h = ut + 1/2 gt^2

u = 0 m/s because the brick was dropped from a height.

h =  1/2 gt^2

h = 1/2 × 10 × (4)^2

h = 80 m

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Answer:

25 atoms

Explanation:

Answer:

A divergent boundary occurs when two tectonic plates move away from each other. Along these boundaries, earthquakes are common and magma (molten rock) rises from the Earth's mantle to the surface, solidifying to create new oceanic crust. When two plates come together, it is known as a convergent boundary.

Explanation:

Answer:

The cat's speed when it slid off the table was 2.53 m/s

Explanation:

Horizontal Motion

When an object is thrown horizontally at a speed v, from a height h, it describes a curved path ruled by gravity, until it hits the ground.

the range or maximum horizontal distance traveled by the object can be calculated as follows:

[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]

The situation described in the problem fits into the concept of a horizontal launch when the cat slides off the table and gets to the ground some distance ahead.

It's given the range as d=1.3m and the height h=1.3 m, thus we can find the cat's speed by solving the equation for v:

[tex]\displaystyle v=d\cdot\sqrt{\frac {g}{2h}}[/tex]

Where [tex]g=9.81\ m/s^2[/tex]

Substituting values:

[tex]\displaystyle v=1.3\cdot\sqrt{\frac {9.81}{2\cdot 1.3}}[/tex]

[tex]\displaystyle v=1.3\cdot\sqrt{\frac {9.81}{2.6}}[/tex]

Calculating:

v = 2.53 m/s

The cat's speed when it slid off the table was 2.53 m/s

Answer:

c) an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity.

Explanation:

When the boy throws the ball upward, there are two forces acting on the ball. First one is the force of gravity, which is constant and equal tot he weight of the ball, acting down. The second force is the force with which the boy throws the ball. This force acts upward and decreases as the ball moves up. This force becomes zero when the ball reaches highest point.

While, on its way down the ball has only one force acting on it. That force is the constant force of gravity acting downward. This gravity force is equal to the weight of the ball.

Hence, the correct option is:

c) an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity.

The force acting on the ball is an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity. Thus the option C is the correct option.

Given,

Force exerted by the air is negligible.

Now the forces acting on the ball after the realese,

            [tex]W=mg[/tex]

Now after reaching the highest point, the ball starts to come down. In this case, only one force will act on it which is the gravity force and this is equal to the weight of the body.

Hence, the force acting on the ball is an almost constant downward force of gravity along with an upward force that steadily decreases until the ball reaches its highest point; on the way down there is only a constant downward force of gravity. Thus the option C is the correct option.

For more about the gravity follow the link below-

https://brainly.com/question/1479537

Answer:

Curve in spine can effect his or her other body parts too in addition to spine. The neck or head can be tilted forward, shoulder could be hunched, hip bone can be uneven, ribs can be effected leading to disorder of lung.

Explanation:

Answer:

1 hour

Explanation:

The tailwind usually adds to the speed of airplanes while the headwind normally decreases the speed of airplanes.

For the initial journey of 200 mph with a 40 mph tailwind speed for 480 miles:

Total speed = 200 + 40 = 240 mph

Also,

Speed = distance/time

time = 480/240 = 2 hours

For the return journey of 200 mph with a 40 mph headwind for 480 miles:

Total speed = 200 - 40 = 160 mph

time = 480/160 = 3 hours

The initial flight took 2 hours while the flight back home took 3 hours. Hence it took 1 hour longer to get back home

Answer:

true

Explanation:

just took the test

Answer:

True

Explanation:

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m​

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

[tex]t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085[/tex]

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ  and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m​

Answer:

The distance between the tip of the shadow and flagpole is 40.80 m.

Explanation:

Let the flagpole be represented by AB and the shadow BC . We have to find the distance between C and A .

Considering the right angle CA is the hypotenuse .

Using the Pythagoras theorem

(Hypotenuse)² = (base)² + (perpendicular)²

CA= √(24)² + (33)²

CA= √576+1089

CA= 40.80 m

The distance between the tip of the shadow and flagpole is 40.80 m.

True. The force of the car on the bug will be equal to the force of the bug on the car.

This statement is true according to Newton's third law of motion.

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction.

In this case, the action is the force of the car on the bug, and the reaction is the force of the bug on the car.

Therefore, the force of the car on the bug is equal in magnitude and opposite in direction to the force of the bug on the car.

More on Newton's law can be found here: https://brainly.com/question/15280051

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Answer:

South = 1.5m

West =4.2m

Explanation:

Kindly see attached a rough draft of the situation

Step one

Given data

From the sketch the direction of the player is along the resultant of the triangle, corresponding to the Hypotenuse

Step two:

Hence for an opponent to tackle him towards the south, he must be at

sin θ= opp/hyp

sin 20=x/4.5

x=sin 20*4.5

x=0.342*4.5

x= 1.5m

Also, for an opponent to tackle him towards the south, he must be at

cos θ= adj/hyp

cos 20=y/4.5

y=cos 20*4.5

y=0.93*4.5

y= 4.2m

Answer:

Yes

Explanation:

Its a good game.

Answer:

oof

Explanation:

Answer and Explanation:

Using a = F/m, we get:

Acceleration for Road A = 480/62 = ~7.74 m/s^2

Acceleration for Road B = 240/62 = ~3.87 m/s^2

Acceleration for Road C = 600/62 = ~9.68 m/s^2

If the net force becomes higher than 600N and the mass stays constant, the acceleration will increase proportionally to how much the force increases and the acceleration will be greater than the acceleration when the force is 600N.

I hope this helps! :)

Answer:

C.

Explanation:

the earth has a high greavitational pull

Answer: C the person and earth

Explanation: I just answered it on a p e x

Answer:

t = 7 seconds

Explanation:

Given that,

The acceleration of a car, a = 4 m/s²

The initial velocity of the car, u = 0

Final velocity of the car, v = 28 m/s

We need to find how much time the car to reach a speed of 28 m/s. Using the definition of the acceleration of a car. It is given by :

[tex]a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{28}{4}\\\\t=7\ s[/tex]

So, it will take 7 seconds to reach a speed of 28 m/s.

Answer:

i can`t help without showing me the choses

Explanation:

Answer:

[tex]Force = 13.8N[/tex]

Explanation:

Given

[tex]Mass = 0.6kg[/tex]

[tex]Acceleration = 23m/s^2[/tex]

Required

Determine the applied force

From the question, we understand that no other force acts on the ball.

i.e. the only applied force on the ball is the force applied by the striker.

So, we apply Newton's second law to solve this question.

And this implies that:

[tex]Force = Mass * Acceleration[/tex]

[tex]Force = 0.6kg * 23m/s^2[/tex]

[tex]Force = 13.8N[/tex]

Hence, the applied force by the striker on the ball is 13.8N