All categories

2 years ago

A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.

Physics

2 answers:

fomenos2 years ago

8 0

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force $N= mg cos 22$

$\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N$

liubo4ka [24]2 years ago

7 0

Answer

580

Explanation

The gravitational force acts perpendicularly downwards (towards the center of the earth).

The normal force of the box is perpendicular to the slanting ramp. This force we are given 538 N.

The gravitational force is the component of this 538 N.

∴ Cos22° = 538/Y Where y is the gravitational force.

Cos22° = 538/Y

Y× Cos22° = 538

Y = 538/Cos22°

= 580.25

Answer to the nearest whole number = 580.

You might be interested in

Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for

g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0

2 years ago

How many significant figures do each of the following numbers have: (a) 214, (b) 81.60, (c) 7.03, (d) 0.03, (e) 0.0086, (f) 3236

Korolek [52]

In determining the number of significant figures in a given number, there are three rules to always remember / follow:

First: All integers except zero are always significant.

Second: Any zeros located between non zeroes are always significant.

Third: A zero located after a non zero in a decimal is always significant whether it is before or after the decimal

Therefore using this rule, the number of significant digits in the given numbers are:

(a) 214 = 3

(b) 81.60 = 4

(d) 0.03 = 1

(e) 0.0086 = 2

(f) 3236 = 4

(g) 8700 = 2

4 0

2 years ago

A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th

alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

$F=mg$

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

$a_y = g = -9.8 m/s^2$

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

$v_x = v_0 cos \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s$

D) 9.3 m/s

The y-component of the shot's velocity is given by:

$v_y = v_0 sin \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s$

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

$v_x = 7.6 m/s$