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2 years ago

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HELP ME PLZ A student wishes to conduct a controlled experiment on the effects of gender on the ability to adapt to left–right i

nversion. Which of the following variables should be the same in all experiments? The test used. The word used. The approximate age of all subjects. All of the above. HELP PLZ

2 answers:

Kruka [31]2 years ago

7 0

Answer:

All of the above lol!

Explanation:

Brainliest Please!

vichka [17]2 years ago

3 0

Answer:

hope i helped even tho i am 100 years late lol not really but STILL

Explanation:

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1.) A projectile is launched at 15 degrees. The landing height is the same as the launch position. At what other angle can the p

allsm [11]

1) The general equations of motion of the projectile on the x and y axis are:
$x(t) = v_0 \cos \alpha t$
$y(t)=v_0 \sin \alpha t - \frac{1}{2}gt^2$
where v0 is the initial velocity, $\alpha$ is the angle with respect to the ground, and $g=9.81 m/s^2$ is the gravitational acceleration. We can see that the motion of the projectile is an uniform motion on the x-axis and an uniformly accelerated motion on the y-axis.

First, we need to find what is the total horizontal displacement of the projectile when it is launched with an angle of $15^{\circ}$ . To do that, we need to find first the time t at which the projectile lands to the ground, and we can find it by requiring y(t)=0:
$v_0 \sin \alpha t - \frac{1}{2}gt^2 =0$
$t( v_0 \sin \alpha - \frac{1}{2} gt)=0$
that has two solutions: t=0 (beginning of the motion) and
$t= \frac{2 v_0 \sin \alpha}{g}$
and this is the time after which the projectile lands to the ground. If we substitute this value into the equation for x(t), we find the total horizontal displacement of the projectile:
$x_1=v_0 \cos \alpha t = v_0 \cos \alpha ( \frac{2 v_0 \sin \alpha }{g} )= \frac{2 v_0^2}{g} \sin \alpha \cos \alpha$
with $\alpha=15^{\circ}$ .

If we call $\beta$ the other angle at which the projectile reaches the same horizontal displacement, the total horizontal displacement in this case is
$x_2 = \frac{2 v_0^2}{g} \sin \beta \cos \beta$
Since the horizontal displacement should be the same in the two cases, we can write x1=x2, which becomes:
$\sin \alpha \cos \alpha = \sin \beta \cos \beta$
Now let's remind that $\cos \theta= \sin (90^{\circ} -\theta)$ so that we can rewrite the equation as
$\sin \alpha \sin (90^{\circ}-\alpha) = \sin \beta \sin (90^{\circ}-\beta)$
and using $\alpha=15^{\circ}$ :
$\sin 15^{\circ} \sin (75^{\circ}) = \sin \beta \sin (90^{\circ}-\beta)$
and we can see that there are two values of $\beta$ that satisfy the equation: $\beta=\alpha=15^{\circ}$ and $\beta=75^{\circ}$ , which is the solution of our problem.

2) The vertical velocity of the ball at the very top of its trajectory is zero. In fact, the very top of the trajectory is the point where the ball starts to go down, so it means it is the moment when the the direction of the vertical velocity of the ball is changing from upward to downward, so it must be the moment when the vertical velocity is zero.

7 0

3 years ago

Water on the kitchen floor at home is considered a safety hazard.

svp [43]

Yes it is, cause someone could slip and accidentally stab themselves

6 0

3 years ago

Read 2 more answers

15m/s is how many Newtons

GuDViN [60]

147.09975 newton meters per second

5 0

3 years ago

What color will the paper appear? Explain why?

Evgen [1.6K]

Answer:

The answer is B

Explanation:

6 0

3 years ago

Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

4 0

3 years ago