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iogann1982 [59]

3 years ago

13

A car is traveling 20 kilometers an hour, then changes direction at a fork in the road. The action that the car takes changes it

s
A.
velocity, but not its speed.

B.
speed, but not its velocity.

C.
acceleration, but not its velocity.

D.
speed, but not its acceleration.

2 answers:

posledela3 years ago

5 0

Answer:

A

Explanation:

Velocity means speed with a direction. Unless the speed changes too, or it has an increase in speed, the answer would be different. Here, the direction changes, making the velocity change.

NISA [10]3 years ago

5 0

It’s ( a )

( velocity, but not its speed)

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Multi celled organisms have a ______ surface area to volume ratio than single celled organisms.

True [87]

A. Larger. It is larger Bc they r all larger than the other

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3 years ago

What includes a distance and a direction? A. Displacement B. Velocity C. Speed D. Acceleration

iren [92.7K]

hi <3

the correct option would be A. displacement. displacement is distance in a direction

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7 0

2 years ago

Read 2 more answers

A person measures his or her heart rate by counting the number of beats in 30s. If 40±1 beats are counted in 30.0±0.5s, what is

nordsb [41]

Answer:

$Rate = 1.33 \pm 0.055$ beats per second

Explanation:

Number of heart beats = $40 \pm 1$

time taken = $30.0 \pm 0.5 s$

now we have

$N = 40 \pm 2.5$ %

$t = 30.0 \pm 1.67$ %

now rate of heart beat is defined as number of heart beat per unit of time

so we have

$Rate = \frac{N}{t}$

$Rate = \frac{40 \pm 2.5}{30 \pm 1.67}$

so we have

$Rate = 1.33 \pm (2.5 + 1.67 )$

$Rate = 1.33 \pm 4.17$ %

$Rate = 1.33 \pm 0.055$ beats per second

7 0

3 years ago

A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕

Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

6 0

3 years ago

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

7 0

3 years ago