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AlladinOne [14]

3 years ago

13

A team of students conducts a series of experiments to investigate collisions. In the first experiment, the two carts collide wi

th each other on a smooth surface. The carts stick together and continue to move forward. In the second experiment, the two carts collide with each other on a rough surface. The carts stick together and quickly come to rest. In both experiments, the initial speeds of the carts are identical. Is there a difference in the total energy of the two experiments?

1 answer:

never [62]3 years ago

7 0

Answer: Yes, there is a difference in the total energy of the two experiments because the first two carts that were on a smooth surface had friction which causes two or more objects to stick together.

Explanation:

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Standing on the roof of a (42.0+A) m tall building, you throw a ball straight up with an initial speed of (14.5+B) m/s. If the b

RSB [31]

First, we must find the vertical distance traveled upwards by the ball due to the throw. For this, we will use the formula:

2as = v² - u²

Because the final velocity v is 0 in such cases

s = -u²/2a; because both u and a are downwards, the negative sign cancels

s = 14.5² / 2*9.81
s = 10.72 meters

Next, to find the time taken to reach the ground, we need the height above the ground. This is:
45 + 10.72 = 55.72 m

We will use the formula
s = ut + 0.5at²

to find the time taken with the initial velocity u = 0.

55.72 = 0.5 * 9.81 * t²

t = 3.37 seconds

4 0

3 years ago

"Two uniform identical solid spherical balls each of mass M and radius R" and moment of inertia about its center 2/5 MR2 are rel

adelina 88 [10]

Answer:

he sphere that uses less time is sphere A

Explanation:

Let's start with ball A, for this let's use the kinematics relations

v² = v₀² - 2g (y-y₀)

indicate that the sphere is released therefore its initial velocity is zero and when it reaches the floor its height is zero y = 0

v² = 0 - 2 g (0- y₀)

v = $\sqrt{2g y_o}$

v = $\sqrt{2 \ 9.8\ H}$

v = 4.427 √H

Now let's work the sphere B, in this case it rolls down a ramp, let's use the conservation of energy

starting point. At the highest point, before you start to move

Em₀ = U = m g y

final point. At the bottom of the ramp

Em_f = K = ½ m v² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

Em₀ = Em_f

mg H = ½ m v² + ½ I w²

angular and linear velocity are related

v = w r

w = v / r

the momentorot of inertia indicates that it is worth

I = $\frac{2}{5}$ m r²

we substitute

m g H = ½ m v² + ½ ( $\frac{2}{5}$ m r²) ( $\frac{v}{r}$ )²

gH = $\frac{1}{2}$ v² + $\frac{1}{5}$ v² = $\frac{7}{10}$ v²

v = $\sqrt{\frac{10}{7} \ g H}$

v = $\sqrt{ \frac{10}{7} \ 9.8 \ H}$

v=3.742 √H

Taking the final speeds of the sphere, let's analyze the distance traveled, sphere A falls into the air, so the distance traveled is H. The ball B rolls in a plane, so the distance (L) traveled can be found with trigonometry

sin θ = H / L

L = H /sin θ

we can see that L> H

In summary, ball A arrives with more speed and travels a shorter distance, therefore it must use a shorter time

Consequently the sphere that uses less time is sphere A

5 0

3 years ago

Jefferson's administration was the beginning of popular government in the United States.

marshall27 [118]

He repealed federal taxes, Jefferson reduced military spending. He bought Louisiana from France. Considering only all these positive points and leaving aside the negative ones, one would say that the statement that "Jefferson's administration was the beginning of popular government in the United States." Is True!

8 0

3 years ago

_____________________ are structures in cardiac muscle tissue that allow impulses to spread across the heart more rapidly.

Darya [45]

Gap junctions in the intercalated discs allow impulses to be spread across the heart more quickly. This is because gap junctions allow particles/signals to pass through, thus making cells with gap junctions more able to interact.

One more thing—you posted this in the physics section rather than biology.

8 0

3 years ago

A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c

tatuchka [14]

We have that the values for F north, F east, F up are

$F_N=1.09090909*10^{-5}$
$F_E=5.18181818*10^{-6}$
$F_E=2*10^{-6}$

From the Question we are told that

electric force $F_1 = 1.2 x 10^{-3} N(N)$

electric force , $F_2=5.7 x 10^{-4} N(E)$

electric force , $F_3=2.2 x 10^{-4} N (U)$

charge on this ball one $q_1= 110 nC.$

charge on this ball two $q_2= -50 nC.$

Generally the equation for the F north is mathematically given as

$F_N=\frac{F_1}{q_1}\\\\F_N=\frac{ 1.2 * 10^{-3} )}{110}$

$F_N=1.09090909*10^{-5}$

For F East

$F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}$

$F_E=5.18181818*10^{-6}$

For F UP

$F_U=\frac{F_3}{q_1}\\\\F_U=\frac{2.2 x 10^-4 }{110}$

$F_E=2*10^{-6}$

For more information on this visit

brainly.com/question/21811998

5 0

3 years ago