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4 years ago

5

A 150-if object takes 1.5 minutes to travel s 2,500-meter straight path. It begins the trip traveling 120 m/s and decelerates to

a velocity of 20 m/s. What was it's acceleration?

1 answer:

aksik [14]4 years ago

4 0

The Answer is it is -1.11 m/s^2.

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Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s

AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J$

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0

4 years ago

Select the correct answer.

Marianna [84]

Answer:

A

Explanation:

According to Newton's second law, acceleration is directly proportional to the net force. As the acceleration increases (when mass is constant), the net force increases. This is represented in the following formula.

$f = ma$

$a = \frac{f}{m}$

a = acceleration

f = force

m = mass

6 0

4 years ago

Choir members are traditionally grouped into four main vocal registers: bass,

Bond [772]

Answer:

The bass has the lowest frequency ,it only causes vibrations but at a lower frequency

7 0

3 years ago

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

6 0

4 years ago

A 4.00-kg particle moves along the x axis. Its position varies with time according to x 5 t 1 2.0t 3, where x is in meters and t

White raven [17]

Answer:

Explanation:

Given equation is ,

x =t + 2 t³ ,

dx/dt = velocity ( v ) = 1 + 6 t²

a) kinetic energy = 1/2 m v² = .5 x 4 ( 1 + 6 t² )² = 2 ( 1 + 6 t²)²

b ) Acceleration = dv /dt = 12 t .

force( F ) = mass x acceleration = 4 x 12 t = 48 t

Power = force x velocity = 48 t x ( 1 + 6 t²). = 48 t + 288 t³ )

work done = ∫ F dx =∫ 48 t x( 1 + 6t² )dt ; = [48t²/2 + 48 x 6 x t³ /3 = 24 t² + 96 t³ )]₀² = 864 J

6 0

3 years ago

Read 2 more answers