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Harlamova29_29 [7]
2 years ago
10

6th grade science I mark as brainliest.​

Physics
1 answer:
DerKrebs [107]2 years ago
3 0

Answer:

2m 13\frac{1}{3}s

Explanation:

1.5m = 1s

200m = \frac{200}{1.5} × 1s

          = 133\frac{1}{3}s

          = 2m 13\frac{1}{3}s

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kenny6666 [7]

Answer:

Explanation:

White dwarfs

6 0
3 years ago
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What is the displacement of the car between t=1s and t=4s
tensa zangetsu [6.8K]

Answer:

Option C. 30 m

Explanation:

From the graph given in the question above,

At t = 1 s,

The displacement of the car is 10 m

At t = 4 s

The displacement of the car is 40 m

Thus, we can simply calculate the displacement of the car between t = 1 and t = 4 by calculating the difference in the displacement at the various time. This is illustrated below:

Displacement at t = 1 s (d1) = 10 m

Displacement at t= 4 s (d2) = 40

Displacement between t = 1 and t = 4 (ΔD) =?

ΔD = d2 – d1

ΔD = 40 – 10

ΔD = 30 m.

Therefore, the displacement of the car between t = 1 and t = 4 is 30 m.

4 0
3 years ago
Which region of the brain serves as the major relay station for most sensory impulses that reach the primary sensory areas of th
Ostrovityanka [42]

Answer:

Thalamus

Explanation:

Thalamus is a mass of gray matter located just above the brain stem, between the cerebral cortex and the midbrain. It plays a role in pain sensation, attention and alertness. It consists of four parts: the hypothalamus, the epythalamus, the ventral thalamus and the dorsal thalamus.

It (Thalamus) serves as the major relay station for most sensory impulses that reach the primary sensory areas of the cerebral cortex from the spinal cord and brain stem.

6 0
3 years ago
Need help with this please
lutik1710 [3]

Explanation:

this help you this is your answer

8 0
2 years ago
A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate area of 25.0 cm2. The plates are charged to a po
Archy [21]

Answer:

(a) Since net charge remains same,after immersion Q is same

(b) I. 14.56pF ii. 3.05V

(c) ΔU = 5.204nJ

Explanation:

a)

C = kεA/d

k=1 for air

ε is 8.85x10-12F/m

A = .0025m2

d = .125m

C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF

Q = CV = .177pF * 244V = 43.188pC

Since net charge remains same,after immersion Q is same

b)

C = kεA/d, for distilled water k is approx. 80

Cwater = Cair x k

=0.177pF x 80 = 14.16pF

Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V

c) Change in energy: ΔU = Uwater - Uair

Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ

Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ

ΔU = 5.204nJ

6 0
3 years ago
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