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Yuliya22 [10]
1 year ago
8

A 2-newton force applied to a 2-kilogram object caused it to move in a straight line 2 meters during an interval of 2 seconds. T

he object gains kinetic energy K during this interval. In which of the following cases will the object gain the same kinetic energy K? A. The same force is applied to a 4-kilogram object for the same time. B. The same force is applied to a 4-kilogram object for the same distance.C. A 4-Newton force is applied to a 4-kilogram object for the same time.D. A 4-Newton force is applied to a 4-kilogram object for the same distance.E. The same force is applied to a 4-kilogram object for 4 seconds.
Physics
1 answer:
yuradex [85]1 year ago
8 0

By the kinetic energy theorem we know that:

W=\Delta K

where W is the work done by the force and delta K is the change in kinetic energy.

In the original scenario the work done by the force is:

W=Fd=(2)(2)=4

Which means that the object gained 4 J in kinetic energy.

Now we need to determine in which scenario given the work done is the same, from the options we notice that in option B we apply the same force for the same distance, which means that the work will be the same and hence the object will gain the same kinetic energy (note that this does not mean that the final velocity will be the same, just that the object will gain the same amount of kinetic energy). In all the other cases we can't conclude the same.

Therefore, the answer is option B.

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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
3 years ago
A plane passes over Point A with a velocity of 8000 m/s north. Forty seconds later it passes over Point B at a velocity of 10,00
Airida [17]

Answer:

The planes’ acceleration from A to B is 500m/s^2

Explanation:

Given that the initial velocity u is 8000m/s

and also given the final velocity v=10,000 m/s

the time taken to move from A to B = 40 second

The acceleration is defined as the rate of change of velocity with time

we know that the expression for acceleration is given as

a=(v-u)/t

substituting our given data into the expression for a we have

a=(10000-8000)/40

a=2000/40

a=500m/s^2

The planes’ acceleration from A to B is 500m/s^2

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