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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.

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Answer: Hello your question is incomplete attached below is the complete question

Ra = 1132 N

Rb = 522.6 N

Pa = 679.7 N

Explanation:

To determine the scalar components Ra

$\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}$

therefore : Ra = $\frac{sin120^o * 750}{sin 35^o}$ = 1132 N

To determine the scalar component Rb

$\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}$

therefore : Rb = $\frac{sin 25^o * 750}{sin 35^o}$ = 522.6 N

To determine the orthogonal projection Pa of R onto

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2 years ago

A uniformly accelerated car passes three equally spaced traffic signs. The signs are separated by a distance d = 25 m. The car p

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Answer:

a) $v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(2(\frac{25}{3})-\frac{25}{3} )m}{3.9s-1.3s} =3.2051 \frac{m}{s}$

b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

Explanation:

The knowable variables are

$d_{t}=25m$

$t_{1}=1.3 s$

$t_{2}=3.9 s$

$t_{3}=5.5 s$

Since the three traffic signs are equally spaced, the distance between each sign is $\frac{25}{3} m$

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b) $v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(25-2(\frac{25}{3}) )m}{5.5s-3.9s} =5.2083 \frac{m}{s}$

Since we know the velocity in two points and the time the car takes to pass the traffic signs

c) $a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} } =\frac{5.2083m/s-3.2051m/s}{5.5s-3.9s} =1.252 \frac{m}{s^{2} }$

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2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

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In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

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m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

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2 years ago

Tyrel is learning about a certain kind of metal used to make satellites. He learns that infrared light is absorbed by the metal,

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It depends on 3 factors:

1. The temperature

2. The specific heat capacity of the metal

3. The thermal conductivity of the metal.

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When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer.

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2 years ago

A construction worker accidentally drops a brick from a high scaffold. a. What is the brick's velocity after 4.0 s? b. How far d

AlekseyPX

Answer:

A. 39.2 m/s

B. 78.4 m

Explanation:

Data obtained from the question include:

Time (t) = 4 s

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A. Determination of the brick's velocity.

Time (t) = 4 s

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = gt

v = 4 × 9.8

v = 39.2 m/s

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B. Determination of how far the brick fall in 4 s.

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h = ½gt²

h = ½ × 9.8 × 4²

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Thus, the brick fall 78.4 m during the time.

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1 year ago