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:

V125BC [204]3 years ago

8 0

D. 128.25 because force=mass x acceleration

gtnhenbr [62]3 years ago

8 0

Answer:

D. 128.25 N

Explanation:

The force exerted on an object and the acceleration of the object are related by the following equation (Newton's second law):

$F=ma$

where

F is the force exerted on the object

m is the mass of the object

a is the acceleration

In this problem, we know:

- The mass of the object: m = 13.5 kg

- The acceleration: a = 9.5 m/s^2

Therefore, we can find the force required to give the object this acceleration:

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

You might be interested in

Which of the waves has the smallest amplitude?

balandron [24]

Answer:

I think its the Blue wave, im not sure so dont take my word for it.

Explanation:

3 0

3 years ago

Hello guys! Can u please help me with physics. I translated it in English. Can yall help me please how much u can!!

DedPeter [7]

1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula

v² - u² = 2a ∆x

where

u = initial speed

v = final speed

a = acceleration

∆x = displacement

At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

0² - (20 m/s)² = 2 (-9.8 m/s²) y

Solve for y :

y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m

2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.

3. At any time t ≥ 0, the body's vertical velocity is given by

v = 20 m/s - gt

At the highest point, we have

0 = 20 m/s - (9.8 m/s²) t

and solving for t gives

t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

4. The body's height y above the ground at any time t ≥ 0 is given by

y = 60 m + (20 m/s) t - 1/2 gt²

Solve for t when y = 0 :

0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²

Using the quadratic formula,

t = (-b + √(b² - 4ac)) / (2a)

(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with

t ≈ 6.09 s

5. At the time found in (4), the body's velocity is

v = 20 m/s - g (6.09 s) ≈ -39.7 m/s

Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.

6 0

3 years ago

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0

3 years ago

I need so much help 50 points!!!!!! complete the graph

motikmotik

First Symbol : Cobalt (Co)

Its Group Number - 9

Its Period Number - 4

Its Family Name - Transition Metal

Second Symbol : Silicon (Si)

Its Group Number - 14

Its Period Number - 2

Its Family Name - Semiconductor

Third Symbol : Astatine (At)

Its Group Number - 17

Its Period Number - 6

Its Family Name - Halogen

Fourth Symbol : Magnesium (Mg)

Its Group Number - 2

Its Period Number - 3

Its Family Name - Alkaline Earth Metal

Fifth Symbol : Xenon (Xe)

Its Group Number - 18

Its Period Number - 5

Its Family Name - Noble Gas

6 0

3 years ago

Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work?

bagirrra123 [75]

Reduce friction because friction just makes things harder

5 0

3 years ago

Read 2 more answers