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3 years ago

Which statements best describe measuring with a graduated cylinder? all that apply.

Physics

2 answers:

melisa1 [442]3 years ago

8 0

My best guess would be C

Maksim231197 [3]3 years ago

8 0

Answer:

a,c,e and f

Explanation:

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Please do this . will mark as brainlest;)

Vika [28.1K]

Answer:

no, not all three contains iron

no, not all three also contain magnets

Explanation:

a has the magnet

b and c have pieces of iron

4 0

3 years ago

Need help anybody thanks

mars1129 [50]

If there are no windows, then the illumination in the room depends
only on the room and what's in it, not on anything outside.
The time of day has no effect. The other things all do.

4 0

3 years ago

A student tested a variety of interacting forces to determine how they would result in motion of an object. If the student used

agasfer [191]

Answer:

D) The variable shown by letter C would result in a movement of the object to the right.

Explanation:

7 0

3 years ago

miniature spring-loaded, radio-controlled gun is mounted on an air puck. The gun's bullet has a mass of 5.00 g, and the gun and

Tamiku [17]

Answer:

12 m/s

Explanation:

From the question,

Applying the law of conservation of momentum,

total momentum before collision = Total momentum after collision

mu+Mu' = mv+Mv'........................... Equation 1

Where m = mass of the bullet, u = initial velocity of the bullet, M = combined mass of the gun and the puck, u' = initial velocity of the gun and the puck, v = final velocity of the bullet, v' = final velocity of the gun and the puck

make v the subeject of the equation

v = [(mu+Mu')-Mv']/m................. Equation 2

Given: m = 5.00 g = 0.005 kg, M = 120 g = 0.12 kg, u = u' = 0 m/s (at rest), v' = 0.5 m/s

Substitute these values into equation 2

v = [0-(0.12×0.5)]/0.005

v = -0.06/0.005

v = -12 m/s

The negative sign can be ignored since we are looking for the speed, which has only magnitude.

Hence the speed of the bullet is 12 m/s

5 0

3 years ago

Starting from rest and moving in a straight line, a cheetah

UNO [17]

The average acceleration of the cheetah is 7.75 m/s²

The given parameters;

velocity of the cheetah, v = 31 m/s

time of motion, t = 4 seconds

The average acceleration of the cheetah is calculated as follows;

$Average \ acceleration = \frac{\Delta velocity}{\Delta time } \\\\Average \ acceleration = \frac{v-u}{t}$

where;

v is the final velocity

u is the initial velocity

t is the time of motion

Assuming the cheetah started from rest, the initial velocity, u = 0

The average acceleration is calculated as follows;

$Average \ acceleration = \frac{v-u}{t} = \frac{31-0}{4} = 7.75 \ m/s^2$

Thus, the average acceleration of the cheetah is 7.75 m/s²

Learn more here: brainly.com/question/17280180

3 0

3 years ago