All categories

3 years ago

In an experiment, a researcher can make claims about causation

1 answer:

8 0

In an experiment, a researcher can make claims about causation if the independent variable changes because of changes made to the dependent variable. Causation works on cause and effect, so the changed independent variable is the cause and the changed dependent variable is the effect. In an experiment the independent variable is changed to determine the dependent variables value, so the two are directly related.

You might be interested in

5. If one object has a greater speed than a second object. does the first necessarily have a greater acceleration? Explain, usin

Sveta_85 [38]

Answer:

Explanation:

5. not necessarily so that the first object could have left with initial velocity and the second not, so even if the second has a greater acceleration its velocity is less than that of the first

6. The acceleration of the motorcycle is

SI System Reductions

Vo = 80 km / h (1000m / 1km) (1h / 3600s) = 22.2 m / s

Vf = 90 km / h (1000m / 1km) (1h / 3600s) = 25 m / s

Vf = Vo + at at = Vf-Vo

am = (Vf-Vo) / t

am = (25 -22.2) / t = 2.8 / t

am= 2.8/t

For the bike we have

Vf = 10 km / h (1000m / 1km) (1h / 3600s) = 2.78 m / s

Vo = 0

ab = (Vf -Vo) / t

ab = (2.78 -0) / t

ab = 2.8/t

Since time is the same for both of us, if we round to Significant figures the two accelerations are equal

7. If when an object is slowing or slowing down.

For example, a car goes north and must stop at the traffic light, the acceleration of the brakes goes south

8. Yes, since an object can go to the left and the acceleration to the right, but the object will lose speed over time

9. in the launch of projectiles the acceleration is negative and the speed after half the path is also negative

10. Car B must be moving to car A, because if they leave together B has more acceleration, bone that travels the distance at the same time

11. When we have friction, the velocity of an object increases by an external force, but the friction also increases the acceleration, but since it is positive, the velocity increases until the acceleration is zero and hence the velocity remains constant.

8 0

3 years ago

What is the impulse of a 2 kg object that starts at 10 m/s and comes to a stop over 0.4 seconds?

gizmo_the_mogwai [7]

Answer: Impulse = 20 Ns

Explanation:

Impulse is the product of force and time

Also impulse = momentum

Where momentum is the product of mass and velocity.

Given that

M = 2kg

V = 10 m/s

Impulse = MV = 2 × 10 = 20 Ns

8 0

2 years ago

A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t

notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is $1.37 m/s^2$

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, $F_f$ , backward

The frictional force can be written as

$F_f = \mu mg$

where

$\mu=0.03$ is the coefficient of kinetic friction

$m=150 kg$ is the mass of the motorbike

$g=9.8 m/s^2$ is the acceleration of gravity

Therefore the net force is given by

$\sum F = F - F_f = F - \mu mg$

And substituting, we find

$\sum F=250 - (0.03)(150)(9.8)=205.9 N$

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

$\sum F = ma$

where

m is the mass

a is the acceleration

In this problem, we have

$\sum F = 205.9 N$ is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

$a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2$

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

$a=1.37 m/s^2$

s = 350 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s$

For this part of the problem, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

$a=1.37 m/s^2$

Solving for t, we find

$t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s$

Learn more about Newton's laws of motion and accelerated motion:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

6 0

3 years ago

A golf ball is struck at 75 m/s at an angle of 60 degrees. What is the

Gnesinka [82]

Answer:

$v_x=37.50$

Explanation:

The rectangular components of a vector $\vec v$ having a magnitude v and angle θ are:

$v_x=v\cos\theta$

$v_y=v\sin\theta$

The golf ball has an initial speed of 75 m/s at an angle of 60 degrees.

The variables of the equations have the values:

v = 75 m/s

θ = 60°

Substituting into the formula:

$v_x=75\cos 60^\circ$

$v_x=75\cdot 0.5$

$v_x=37.5~m/s$

Without specifying units and with precision to the hundredths place:

$\mathbf{v_x=37.50}$

7 0

3 years ago

Describe briefly what specific information is given by each of the four quantum numbers

____ [38]

The following information is given by each four quantum numbers
Main energy level,
Sublevels in main energy levels,
Number of orbitals per sublevel,
number of orbitals per main energy level, number of electrons per sublevels
hope it helps

3 0

3 years ago