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3 years ago

Kid A has jumped off the sled! (RIP)

Physics

1 answer:

balandron [24]3 years ago

5 0

Answer:

46.45 m/s

Explanation:

Total momentum before jump = Total momentum after jump

11.7 * ( 36.7 + 45.2 ) = ( 36.7 * (-31.1) ) + ( 45.2 * v )

v*45.2 - 1141.37 = 958.23

v = 2099.6/45.2 = 46.45

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The energy attributed to an object by virtue of its motion is known as _______________. potential energy mass-energy radiative e

Crazy boy [7]

I think the answer is :
Kinetic energy.
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2 years ago

which of the numbers on this figure indicates typical continental conditions (regional metamorphism)?

Tpy6a [65]

The number that indicates typical continental conditions (regional metamorphism) is that showing schist and gneiss rocks.

<h3>What is regional metamorphism?</h3>

Regional metamorphism occurs when rocks undergoes changes as a result of high temperatures and pressure deep within the earth's crust.

Regional metamorphic rocks are usually foliated or squashed in appearance.

Examples of regional metamorphism rocks are schist and gneiss rocks.

Therefore, the figure that indicates typical continental conditions (regional metamorphism) is that showing schist and gneiss rocks.

Learn more about regional metamorphism at: brainly.com/question/14678538

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4 0

2 years ago

Read 2 more answers

A 0.141 kg pinewood derby car is moving 1.33 m/s . What is its momentum?

kompoz [17]

momentum= mass × velocity = 0.141kg×1.33m/s= 0.18753kg m/s = 0.188kg m/s (3s.f.)

7 0

3 years ago

There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C

liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

$V_{tot} = V_{q1} + V_{q2}$

$V_q = \dfrac{k \cdot q}{r}$

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

$V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0$

The electric field at the point exactly midway between the plates, $V_{tot}$ = 0

3) The electric field, 'E', between plates is given as follows;

$E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C$

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, $F_e$ = E × e

∴ $F_e$ = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, $F_e$ ≈ 1.807 × 10⁻⁷ N

4 0

2 years ago

Sam, whose mass is 60 Kg, is riding on a 5.0 kg sled initially traveling at 8.0 m/s. He

umka2103 [35]

<h3>Answer: 130 newtons</h3>

===============================================================

Explanation:

We'll need the acceleration first.

The initial speed (let's call that Vi) is 8.0 m/s
The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
This happens over a duration of t = 4.0 seconds

The acceleration is equal to the change in speed over change in time

a = acceleration

a = (change in speed)/(change in time)

a = (Vf - Vi)/(4 seconds)

a = (0 - 8.0)/4

a = -8/4

a = -2

The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.

Here's a further break down of Sam's speeds at the four points of interest

At 0 seconds, he's going 8 m/s
At the 1 second mark, he's slowing down to 8-2 = 6 m/s
At the 2 second mark, he's now at 6-2 = 4 m/s
At the 3 second mark, he's at 4-2 = 2 m/s
Finally, at the 4 second mark, he's at 2-2 = 0 m/s

Next, we'll apply Newton's Second Law of motion

F = m*a

where,

F = force applied
m = mass
a = acceleration

We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg

So the force applied must be:

F = m*a

F = 65*(-2)

F = -130 newtons

This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.

The magnitude of this force is |F| = |-130| = 130 newtons

8 0

3 years ago