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4 years ago

8

The speed of a boat undergoing constant acceleration increases from 5 meters per second to 15 meters per second over a period of

20 seconds. How far does the boat travel during this time?

1 answer:

defon4 years ago

4 0

Should be 250m. check with your teacher and let me know

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If a stone is dropped from a height of 400 feet, its height after t seconds is given by s = 400 − 16t2. Find its instantaneous v

soldier1979 [14.2K]

Answer:

$v = -32 t$

Explanation:

given,

s = 400- 16 t²

we know,

Velocity of an object is defined as the change in displacement per unit change in time.

velocity an also be return as

$v = \dfrac{ds}{dt}$

$v = \dfrac{d}{dt}(400-16t^2)$

$v= 0 -2\times 16 t$

$v = -32 t$

Hence, instantaneous velocity function given by $v = -32 t$

To calculate instantaneous velocity, you need to insert value of time.

ex, instantaneous velocity at t = 4 s

v = -32 x 4 = -128 m/s.

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3 years ago

The generation of a magnetic field by an electric current is___ .

Alex787 [66]

Answer:

Electromagnetic induction

Explanation:

The process of generating electric current with a magnetic field. It occurs whenever a magnetic field and an electric conductor move relative to one another so the conductor crosses lines of force in the magnetic field.

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2 years ago

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance

xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

$\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}$

$\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}$

$v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}$

$v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}$

$v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}$

v₂ = 7.6 x 10⁴ m/s

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3 years ago

Each tyre of a car has an area of 100 cm2 in contact with the ground. The car has a mass of 1600 kg. The weight of the car is eq

vlabodo [156]

Answer:

b is the answer

Explanation:

tq friend b is the answer

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3 years ago

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An object moves in the eastward direction at constant speed. A net force directed northward acts on the object for 5.0 s. At the

Pie

Answer: D

The final velocity of the object will be directed north of east

Explanation:

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4 years ago

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