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2 years ago

Can you answer this for me please

Physics

1 answer:

lina2011 [118]2 years ago

8 0

I think it might be independant variable.. but not so sure

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Distance v. Time

Allushta [10]

Answer:

(m)

2 ( s )

Explanation:

ok...........

3 0

2 years ago

What is the likely identity of a metal if a sample has a mass of 63.5 g when measured in air and an apparent mass of 60.2 g when

Gre4nikov [31]

Answer:

Gold

Explanation:

Given:

Mass of sample = 63.5 g

Mass of water = 60.2 g

Find:

Object

Computation:

Mass of water displaced = 63.5 g - 60.2 g

Mass of water displaced = 3.3 g

So, volume in water = 3.3 cm³

Density = Mass / Volume

Density = 63.5 g / 3.3

Density = 19.24

So,

Object ,must be gold.

7 0

2 years ago

A balloon filled with helium gas at 1.00 atm occupies 15.6 L. Will the volume of the balloon increase or decrease in the upper a

Mariana [72]

Answer:

The volume of the balloon increases in the upper atmosphere.

Explanation:

p1= 1 atm

p2= 0.15 atm

V1= 15.6 L

V2= ?

p1*V1= p2 * V2

V2= (p1/p2)*V1

V2= 104 L

6 0

2 years ago

A 5kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N opposing the motion. Draw a fo

Taya2010 [7]

Answer:

Net force on the block is 32 N.

Acceleration of the object is 6.4 m/s².

Explanation:

Let the acceleration of the object be $a$ m/s².

Given:

Mass of the block is, $m=5\ kg$

Force of pull is, $F=40\ N$

Frictional force on the block is, $f=8\ N$

The free body diagram of the object is shown below.

From the figure, the net force in the forward direction is given as:

$F_{net}=F-f=40-8=32\ N$

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

$F_{net}=ma\\32=5a\\a=\frac{32}{5}=6.4\ m/s^2$

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².

3 0

2 years ago

If 3600 j of work is done in 3.0 s what is the power<br>0.00083W<br>1200W<br>3600W<br>11000W

Viktor [21]

Answer:

1200 W

Explanation:

Power is given by the ratio between work done and time taken:

$P=\frac{W}{t}$

where W is the work done and t the time taken.

In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

$P=\frac{3600 J}{3.0 s}=1200 W$

5 0

2 years ago

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