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dybincka [34]
4 years ago
5

A toy car moves around a circular track at constant speed. It suddenly doubles its speed - a change of a factor of 2. As a resul

t, the centripetal acceleration changes by a factor of No change since the radius doesn't change. 1/4. 2. 1/2. 0 4.
Physics
1 answer:
lara [203]4 years ago
5 0

Answer:

4

Explanation:

because

centripital \: force =  \frac{m {v}^{2} }{ r} \\ if \: velocty \: will \: be \: double\\  so \:  \\ force \: will \: be \:4 \: times

plz...

mark it as a brilliant answer

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Parallel circuits are most likely to be used in which of the following?
lara31 [8.8K]
They're most likely to be used in a house because if you turn on a switch in ur house the rest of them stay off they don't turn on
6 0
3 years ago
Read 2 more answers
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
Can somone pls help me??!! i’m very stuck
Alchen [17]

Answer:

its the third one

Explanation:

3 0
3 years ago
What is the speed of a transverse wave in a rope of length 3. 1 m and mass 86 g under a tension of 380 n?
Nitella [24]

117 m/sec is the speed of a transverse wave in a rope of length 3. 1 m and mass 86 g under a tension of 380 n.

The wave speed v is given by

v= √τ/μ

​where τ is the tension in the rope and μ is the linear mass density of the rope.

The linear mass density is the mass per unit length of rope :

μ= m / L = (0.086 kg)/(3.1 m)=0.0277 kg/m.

v= \sqrt{ \frac{380 N}{0.0277 kg/m}}  = 117.125 m/sec (approx. 117 m/sec

In physics, a transverse wave is a wave whose oscillations are perpendicular to the direction of the wave's advance. This is in contrast to a longitudinal wave which travels in the direction of its oscillations. Water waves are an example of transverse wave.

Transverse waves commonly occur in elastic solids due to the shear stress generated; the oscillations in this case are the displacement of the solid particles away from their relaxed position, in directions perpendicular to the propagation of the wave. These displacements correspond to a local shear deformation of the material. Hence a transverse wave of this nature is called a shear wave. Since fluids cannot resist shear forces while at rest, propagation of transverse waves inside the bulk of fluids is not possible.

Learn more about Transverse waves here : brainly.com/question/13761336

#SPJ4

3 0
2 years ago
The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold
Liula [17]

Answer:

L_{o}=0.1224m

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m

6 0
4 years ago
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