All categories

4 years ago

A 19.7 kg sled is pulled with a 42.0 N force at a

Physics

2 answers:

sertanlavr [38]4 years ago

5 0

The normal force on the sled is 190.9 N.

Explanation:

For this problem, we have to analyze the forces acting along the vertical direction on the sled.

We have:

- The weight of the sled, W = mg, acting downward, where

m = 19.7 kg is the mass of the sled

g = 9.8 m/s^2 is the acceleration of gravity

- The normal force on the sled, N, acting upward

- The vertical component of the pulling force, $F sin \theta$ , acting upward, where

F = 42.0 N is the magnitude of the force

$\theta=3.0^{\circ}$ is the angle

Since the sled is in equilibrium along the vertical direction, the equation of the force is

$F sin \theta + N - mg = 0$

And solving for N, we find

$N=mg -Fsin \theta = (19.7)(9.8) - (42.0)(sin 3)=190.9 N$

Learn more about forces and weight here:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

ZanzabumX [31]4 years ago

5 0

Answer:

164 N

Explanation:

You might be interested in

1. Find the magnitude of the gravitational force (in N) between a planet with mass 8.00 ✕ 1024 kg and its moon, with mass 2.75 ✕

MrRissso [65]

Answer:

1.358 10e8 have good day please mark brainliest

8 0

3 years ago

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What

Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

$\alpha = 6261 \ rev/minutes^2$

$\theta = 613 \ revolutions$

Explanation:

From the question we are told that

The initial angular speed is $w_i = 1120 \ rev/minutes$

The angular speed after $t = 13.8 s = \frac{13.8}{60 } = 0.23 \ minutes$ is $w_f = 2560 \ rev/minutes$

The time for revolution considered is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$

Generally the angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_i }{t}$

=> $\alpha = \frac{2560 - 1120 }{0.23}$

=> $\alpha = 6261 \ rev/minutes^2$

Generally the number of revolution made is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$ is mathematically represented as

$\theta = \frac{1}{2} * (w_i + w_f)* t$

=> $\theta = \frac{1}{2} * (1120+ 2560 )* 0.333$

=> $\theta = 613 \ revolutions$

5 0

3 years ago

a small negatively charged sphere with a mass of 5.4*10^-5 is suspended between two parallel plates. the potential difference is

labwork [276]

Here, Fe = Fg
q.E = m.g
We have: E = 360 V
m = 5.4 × 10⁻⁵
g = 9.8 m/s² [ constant value for earth system ]

Substitute their values into the expression:
q (360) = 5.4 × 10⁻⁵ × 9.8
q = 52.92 × 10⁻⁵ / 360
q = -1.47 × 10⁻⁶ [ negative sign represents the nature of charge ]

So, Your Final answer would be 1.47 × 10⁻⁶

Hope this helps!

3 0

4 years ago

Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest an

Morgarella [4.7K]

Answer:

2.218 m/s; 4.4375 m/s; 0.01981 m/s²

Explanation:

for more details see the attached picture.

4 0

3 years ago

A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, does the distance to the cat appear to be less tha

lakkis [162]

Answer:

p = -q

he distance is equal to the current distance, so the distance does not change

Explanation:

For this exercise we can solve it using the equation of the constructor

1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

For a flat surface the radius is at infinity, therefore 1 / f = 0, which implies

1 / p = - 1 / q

p = -q

Therefore the distance is equal to the current distance, so the distance does not change

7 0

3 years ago