Convert the following : 6400km into metre

The following time dependent force is applied to a 2 kg block that is initially at rest on a frictionless and horizontal surface

DerKrebs [107]

Answer:

The displacement of the block is 9 m.

Explanation:

Given that,

Mass of block = 2 kg

Time = 3 sec

The force is

$F(t)=4t$

The acceleration is

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{4t}{2}$

$a=2t$

We need to calculate the velocity

Using formula of acceleration

$v=\int_{0}^{t}{a dt}$

Put the value into the formula

$v=\int_{0}^{3}(2t dt)$

$v=(\dfrac{2t^2}{2})_{0}^{3}$

$v=(t^2)_{0}^{3}$

$v=9\ m/s$

We need to calculate the displacement

Using formula of velocity

$s=\int_{0}^{3}(v dt)$

$s=(t^2)_{0}^{3}$

$s=(\dfrac{t^3}{3})_{0}^{3}$

$s=9\ m/s$

Hence, The displacement of the block is 9 m.

8 0

3 years ago

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What is the magnification formula?

erica [24]

Answer:

the height of the image ÷ by the height of the object.

Explanation:

7 0

4 years ago

A vector in the xy plane has components -14.0 units in the x-direction and 30.0 units in the y-direction. What is the magnitude

OverLord2011 [107]

$\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}$

We would calculate the magnitude by applying pythagorean theorem:

$\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}$

$\longrightarrow \sf{Magnitude = 33.12}$

$\longrightarrow \sf{The \: vector \: is \: (- 14, 30)}$

The angle between two vectors is given by the formula:

$\sf{\longrightarrow \small \cos \emptyset = \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }$

In two dimensional, the x axis of vector form is:

$\small\sf{\longrightarrow (b1, b2) = (1, 0) }$

$\sf{\longrightarrow \small \cos \: \emptyset = \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }$

$\small\longrightarrow \sf{ \dfrac{14}{33.12} }$

$\small\longrightarrow \sf{\emptyset \: = arcCos (\dfrac{ - 14}{33.12} )}$

$\small\longrightarrow \sf{\emptyset= 115^\circ}$

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

$\small\bm{The \: angle \: between \: the \: vector \: }$

$\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}$

6 0

2 years ago

BRAINLIEST!!! how does the electric force between two charged particles change if the distance between them is reduced by a fact

Bogdan [553]

Answer:

F = K Q1 * Q2 / R^2 force between 2 charged particles

F2 / F1 = R1^2 / R2^2 = (R1 / R2)^2 = (R1 / R1 /2)^2 = 2^2 = 4

d) the force would be increased by a factor of 4

3 0

2 years ago

An object is dropped from 600 m. If it is initially at rest and achieves a top speed of 45 m/s just as it hits the ground, what

Brums [2.3K]

Answer:

1.68m/s^2

Explanation:

using V^2=U^2+2×a×s formula.

If,

(Final Velocity)V=45

(Initial Velocity)U=0 because it starts from rest

(Distance)S=600m

(acceleration)a= ?

now,

using formula,

45^2=0^2+2×a×600

2025= 1200a

a=2025÷1200

a = 1.68m/s^2(The unit of acceleration is m/s^2)

Therefore the acceleration is 1.68m/s^2

Hope it works !!!

3 0

3 years ago

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