All categories

3 years ago

16. An object has a mass of 13.5 kilograms. What force is required to accelerate it to a rate of 9.5 m/s2?

Physics

2 answers:

Firdavs [7]3 years ago

7 0

Answer: The correct answer is option D.

Explanation:

Mass of an object = 13.5 kg

Rate of the acceleration = $9.5 m/s^2$

$Force=Mass\times Acceleration=ma$ (Newton's second law of motion)

$Force=13.5 kg\times 9.5 m/s^2=128.25 kg m/s^2=128.25 Newtons$

128.25 Newtons of force is required to accelerate it to a rate of $9.5 m/s^$ .

Hence,the correct answer is option D.

Ivahew [28]3 years ago

3 0

We can solve the problem by using Newton's second law of motion:

$F=ma$

where F is the force required to accelerate the object, m is its mass, and a is the acceleration.

In this problem, we are told that the mass of the object is m=13.5 kg, while its acceleration is a=9.5 m/s^2. Substituting these data into the equation, we find the force required to produce these acceleration:

$F=ma=(13.5 kg)(9.5 m/s^2)=128.25 N$

So, the correct answer is

D. 128.25 N

You might be interested in

Part 1: A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the en

Anit [1.1K]

1) c. 2 m/s

Explanation:

The relationship between frequency, wavelength and speed of a wave is

$v=\lambda f$

where

v is the speed

$\lambda$ is the wavelength

f is the frequency

For the wave in this problem,

f = 4 Hz

$\lambda=0.5 m$

So, the speed is

$v=(0.5 m)(4 Hz)=2 m/s$

2) a. 2.8 m/s

The speed of the wave on a string is given

$v=\sqrt{\frac{T}{\mu}}$

where

T is the tension in the string

$\mu$ is the linear mass density

In this problem, we have:

$T=2 \cdot 4 N=8 N$ (final tension in the rope, which is twice the initial tension)

$\mu = 1 kg/m$ --> mass density of the rope

Substituting into the formula, we find

$v=\sqrt{\frac{8 N}{1 kg/m}}=2.8 m/s$

5 0

3 years ago

a coil has n turns enclosing an area of a. in a physics laboratory experiment, the coil is rotated during the time interval δt f

Mrac [35]

The correct answer for the question that is being presented above is this one:

Phi = BAsin(theta)
1. Phi(i) = BA 
2. Phi(f) = 0 
3. EMF = N(phi(i)-phi(f))/deltat

Here are the follow-up questions:

1. What is the total magnitude Phi_initial of the magnetic flux through the coil before it is rotated? 
2. What is the magnitude Phi_final of the total magnetic flux through the coil after it is rotated? 
3. What is the magnitude of the average emf induced in the coil?

8 0

3 years ago

A cat is chasing a mouse across a 1.3 m tall dining table. The mouse darts to the side and the cat accidentally slides off, land

erastovalidia [21]

Answer:

1) 0.51 seconds.

2) 1.45 m/s.

Explanation:

given, height from which cat falls = 1.3 m

we know that, s = ut + $\frac{1}{2}$ at².

here if we consider cat moment only in downward direction,

intial velocity of cat in downward direction , u = 0.

so, time, t = $\sqrt{\frac{2h}{g} }$ .

⇒ t = $\sqrt{\frac{2(1.3)}{9.81} }$ = 0.51 seconds.

t = 0.51 seconds.

now, consider cat moment only in forward direction

s = ut , since acceleration is zero in forward direction

⇒ u = $\frac{s}{t}$ .

so, u = $\frac{0.75}{0.51}$ = 1.45 m/s .

6 0

3 years ago

PLS PLS HELP TIMED TEST!!!!!!

PolarNik [594]

Answer:

Explanation:

hey

6 0

3 years ago

Read 2 more answers

Mercury, a planet in our Solar System, and Callisto, a moon of the planet Jupiter, are about the same size. However, Mercury’s m

TiliK225 [7]

You may jump higher because the more the mass of the planet, the more gravitational force. There is less mass(and gravity) on Callisto so you wouldn’t be weighed down as much and can jump higher. Whereas on Jupiter there is more weight holding you down.

3 0

3 years ago

Read 2 more answers