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koban [17]
3 years ago
11

Leo places a plant in front of the center of curvature of a concave mirror. Which characteristics will the image of the plant ha

ve? Check all that apply.
Physics
2 answers:
photoshop1234 [79]3 years ago
6 0

Answer:

1, 3, 5,

Explanation:

Ket [755]3 years ago
3 0
1,3,5 that is the answer
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Answer the following question according to formal grammatical rules.
VARVARA [1.3K]

Answer:

c

Explanation:

just trying to follow basic grammar.

5 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
3 years ago
(1) Which appliance is designed to transfer electrical energy to kinetic energy?
sweet-ann [11.9K]

Answer:

bb kettle

Explanation:

it transfres electricsl to kinetic

8 0
3 years ago
What will happen when you slide the sponges in different directions?
Ede4ka [16]

Transform boundary

is the answer from stemscopes

4 0
2 years ago
A tennis ball is released from a height of 4.0 m above the floor. After its third bounce off the floor, it reaches a height of 1
diamong [38]

Answer:

The percentage of its mechanical energy does the ball lose with each bounce is 23 %

Explanation:

Given data,

The tennis ball is released from the height, h = 4 m

After the third bounce it reaches height, h' = 183 cm

                                                                       = 1.83 m

The total mechanical energy of the ball is equal to its maximum P.E

                                      E = mgh

                                          = 4 mg

At height h', the P.E becomes

                                      E' = mgh'

                                           = 1.83 mg

The percentage of change in energy the ball retains to its original energy,

                                 \Delta E\%=\frac{1.83mg}{4mg}\times100\%

                                  ΔE % = 45 %

The ball retains only the 45% of its original energy after 3 bounces.

Therefore, the energy retains in each bounce is

                                   ∛ (0.45) = 0.77

The ball retains only the 77% of its original energy.

The energy lost to the floor is,

                                E = 100 - 77

                                   = 23 %

Hence, the percentage of its mechanical energy does the ball lose with each bounce is 23 %      

5 0
3 years ago
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