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3 years ago

Which of the following are appropriate acceleration units? km/hr2 m/s/s ft/s miles/hr/min sec/km/m

Physics

2 answers:

Reika [66]3 years ago

8 0

I think it is m/s/s or miles/hr/min

docker41 [41]3 years ago

6 0

Km/h² or m/s/s are the appropriate units of the acceleration.

Acceleration is the rate change of velocity of an object. it is given by

a=ΔV/Δt

Δv= change in speed

Δt= time taken

Δv is measured in meter of km

Time is measured in second s or hour h

so the units of acceleration are km/h² or m/s/s

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9. A 227 kg object is moved a distance of 2.4 m forward by a force. If 686 J of work is done on the object, what is the object’s

Korolek [52]

1.259ms^2

Explanation:

since, WORK DONE = FORCE*DISTANCE

AND, FORCE=MASS*ACCELERATION

SO, THE WORK DONE BECOMES=MASS*ACCELERATION*DISTANCE

ACCELERATION=WORK/(MASS*DISTANCE)

AND, WORK=686J

MASS=227kg

DISTANCE=2.4m

THEREFORE, ACCELERATION=686/(227*2.4)

=686/544.8

=1.259ms^2

4 0

3 years ago

Calculate the frequency of the red light emitted by a neon sign with a wavelength of 690 nm

Ne4ueva [31]

Answer:

$4.35 \times 10^{14} Hz$

Explanation:

The frequency of a light is inversely proportional to its wavelength. It is given by:

$f = \frac{v}{\lambda}$

The speed of the red light, v = 3.0 × 10⁸ m/s

The wavelength of the red light, λ = 690 nm = 690 ×10⁻⁹ m

$f = \frac{3.0 \times 10^8 m/s}{690\times 10^{-9}m} = 4.35 \times 10^{14} Hz$

Thus, the frequency of red light emitted by neon sign having wavelength 690 nm is $4.35 \times 10^{14} Hz$

7 0

3 years ago

Read 2 more answers

El Nino begins with which of the following?

Rudik [331]

The eastward movement of winds and currents in the pacific

6 0

3 years ago

Read 2 more answers

A block of mass 0.510 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T

maria [59]

Answer:

x=0.46m, speed=7.9m/s

Explanation:

Using the concept of conservation of energy:

1. kinetic energy of mass m and velocity v: $E_k=\frac{1}{2}mv^2$

2. gravitational potential energy of mass m, grav. acc. g and height h: $E_g=mgh$

3. potential energy in a spring with spring constant k and displacement from equilibrium x: $E_s=\frac{1}{2}kx^2$

Calculating x:

$\frac{1}{2}mv_a^2=\frac{1}{2}kx^2$

$x=\sqrt{\frac{m}{k}}v_a$

Calculating the speed:

$\frac{1}{2}mv_a^2 +mgh_a=\frac{1}{2}mv_b^2+mgh_b + W_{friction}$

$h_a=0, h_b=2R,W_{friction}=F_{friction}\times distance=7\pi R$

$\frac{1}{2}mv_a^2=\frac{1}{2}mv_b^2+2mgR+7\pi R$

Solving for $v_b$ :

$v_b=\sqrt{v_a^2-4gR-14\pi\frac{R}{m}}$

7 0

3 years ago

David lifts a book 1.2 meters from the floor to the top of his desk. If the book weighed 0.50 kg, what is the gravitational pote

Anika [276]

Answer:

5.886 J

Explanation:

Given:

The mass of the book is, $m=0.5$ kg

Height of lift is, $\Delta h=1.2$ m

Acceleration due to gravity is, $g=9.81$ m/s²

Now, gain in gravitational potential energy is a function of change in position and is given as:

$\Delta PE=mg\Delta h$

Here, $\Delta PE$ is the change in gravitational potential energy.

Plug in 0.5 kg for $m$ , 9.81 for $g$ and 1.2 for $\Delta h$ . Solve for $\Delta PE$

$\Delta PE=0.50\times 9.81\times 1.2=5.886\ J$

Therefore, the gain in gravitational energy of the book is 5.886 J.

7 0

3 years ago