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Damm [24]
3 years ago
5

A body moves in a straight line. At time, it's acceleration is given by a = 12t + 1. When t =0, the velocity of the body v is 2

metres per second and its displacement from the origin s is 0.
1. Express v and s in terms of t.
2. Determine velocity and displacement of the body after 3 seconds. ​
Physics
1 answer:
madreJ [45]3 years ago
5 0

Answer:

v = 6t² + t + 2, s = 2t³ + ½ t² + 2t

59 m/s, 64.5 m

Explanation:

a = 12t + 1

v = ∫ a dt

v = 6t² + t + C

At t = 0, v = 2.

2 = 6(0)² + (0) + C

2 = C

Therefore, v = 6t² + t + 2.

s = ∫ v dt

s = 2t³ + ½ t² + 2t + C

At t = 0, s = 0.

0 = 2(0)³ + ½ (0)² + 2(0) + C

0 = C

Therefore, s = 2t³ + ½ t² + 2t.

At t = 3:

v = 6(3)² + (3) + 2 = 59

s = 2(3)³ + ½ (3)² + 2(3) = 64.5

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When the number of atoms on the right side a chemical equation matches the number of atoms on the left side of a chemical equati
Hitman42 [59]

Answer:

Balanced.

Explanation:

A Balanced Chemical equation is a scientific term that describes a chemical equation that has the same number of atoms on each side of the equation.

Hence, when the number of atoms on the right side of a chemical equation matches the number of atoms on the left side of a chemical equation, it is said to be BALANCED.

7 0
3 years ago
A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle
tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

-h= vsinA\times t-gt^2/2

putting values h=15 m, v=0.8

-15 = 0.8vt - 4.9t^2  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

-15 = 0.8\times40/0.6 - 4.9t^2

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = -H =v sinA t - gt^2/2

⇒ 4.9t^2 - 8.9\times0.8t - 100 = 0

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0
3 years ago
The desert sand is very hot during the day and very cold at night. What does this tell you about its specific heat capacity?
erma4kov [3.2K]

Answer: The specific heat capacity is very low.

Explanation:

The specific heat capacity of a body is defined as the heat energy required by a body to cause a unit change in its temperature. The value is over low that is why it is easier for the desert sand to easily get very hot during the day. Conversely, it is very easy for the desert sand to lose it's heat a cool breeze pass over it in the night making it very cold in the night. This value also defines how long the desert sand can retain heat. Therefore, the desert sand has a low specific heat capacity.

8 0
3 years ago
A 16.75 g piece of iron absorbs 986.75 joules of heat energy, and it’s temperature changes from 25 degrees Celsius to 175 degree
Anestetic [448]

Answer:

0.39 J/g°c

Explanation:

= heat / unit of mass × unit of temperature

986.75J/16.75g

= 58.9 J/g

∆T=175°c - 25°c = 150°c

986.75 / 150°c = 6.578

986.75 / 16.75g.150°c = 0.30 j/g°c

8 0
3 years ago
A physical quantity X is connected from X = ab2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3
Hoochie [10]

Answer: 11%

Explanation:

Given that

X = ab^2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3 respectively.

Percentage error of b = 2%

Percentage error of b^2 = 2 × 2 = 4

When you are calculating for percentage error that involves multiplication and division, you will always add up the percentage error values.

Percentage error of X will be;

Percentage error of a + percentage error of b^2 + percentage error of c

Substitute for all these values

4 + 4 + 3 = 11%

Therefore, percentage error of X is 11%

3 0
3 years ago
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