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3 years ago

The part of the scientific method where we record data is the

Physics

2 answers:

STatiana [176]3 years ago

8 0

Answer is D) experiment because you have to record step by step

tensa zangetsu [6.8K]3 years ago

3 0

The answer is d. experiment

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A car traveling at a speed of 13 meters per second accelerates uniformly to a speed of 25 meters per second in 5.0 seconds. 11-

Aleks04 [339]

Answer:

$a=2.4\ m/s^2$

Explanation:

Given that,

Initial speed of a car, u = 13 m/s

Final speed of a car, v = 25 m/s

Time, t = 5 s

We need to find the acceleration of the car during this 5.0 second time interval. Let a is the acceleration. It can be calculated as :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{25-13}{5}\\\\=2.4\ m/s^2$

So, the acceleration of the car is $2.4\ m/s^2$ .

7 0

3 years ago

A object travels at constant negative acceleration. What does the graph of the object's velocity as a fun

melomori [17]

Constant = straight line
“Travels at constant negative acc.”
Which is negative slope

Solution: B. Straight line w/ neg. slope

7 0

3 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago

The electric force between two charged objects is influenced by

tresset_1 [31]

<span> attraction between the relative abundance of electrons in one object and protons in the other

</span>

7 0

3 years ago

If vx=9.80 units and vy=-6.40 units, determine the magnitude and direction of v

dexar [7]

The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:

v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units

5 0

3 years ago