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2 years ago

3

A 1.0-kg block is pushed up a rough 22° inclined plane by a force of 7.0 n acting parallel to the incline. the acceleration of t

he block is 1.4 m/s2 up the incline. determine the magnitude of the force of friction acting on the block.

1 answer:

Mama L [17]2 years ago

6 0

<span>The answer is 1.9 N done with the following equation:
7-mgSinG-f=ma
f=7-mgSinG-ma
=1.92N

It's easiest if you draw out the diagram of the problem first and fill in all the known and missing parts to come up with the equation that you will need to use to solve the problem. In this case we know the angle, weight, force, and acceleration and must find the magnitude acting upon the block.</span>

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Dentists' chairs are examples of hydraulic-lift systems. If a chair weighs 1400 N and rests on a piston with a cross-sectional a

NeX [460]

Answer:

Force applied to smaller cross section is

= 82.63 N

Explanation:

As we know

$F_2 A_1 = F_1 A_2$

where $F1, F2$ signifies the weight of the two chair in a hydraulic-lift system

And $A_1, A_2$ signifies the area of the two respective chairs in a hydraulic-lift system

Given -

$F2=1400$ N

$A1 =1220$ Square centimeter

$A_2 = 72$ Square centimeter

Substituting the given values in above equation, we get -

$1400 * 72 = F1 * 1220\\F2 = 82.63$

Force applied to smaller cross section is

= 82.63 N

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2 years ago

A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

v is image distance

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm$

Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3$

So, the magnification of the lens is 3.

3 0

2 years ago

A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of fl

Korvikt [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is $\frac{dT}{dt} = 1.016 ^oC/m$

Explanation:

From the question we are told that

The velocity field with which the bird is flying is $\vec V = (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s$

The temperature of the room is $T(x, y, u) = 400 -0.4y -0.6z-0.2(5 - x)^2 \ ^o C$

The time considered is t = 10 \ seconds

The distance that the bird flew is x = 1 m

Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

$\frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 * (5-x)] [-\frac{dx}{dt} ]$

Here the negative sign in $\frac{dx}{dt}$ is because of the negative sign that is attached to x in the equation

So

$\frac{dT}{dt} = -0.4v_y -0.6v_z -0.2[2 * (5-x)][ -v_x]$

From the given equation of velocity field

$v_x = 0.6x$

$v_y = 0.2t$

$v_z = -1.4$

So

$\frac{dT}{dt} = -0.4[0.2t] -0.6[-1.4] -0.2[2 * (5-x)][ -[0.6x]]$

substituting the given values of x and t

$\frac{dT}{dt} = -0.4[0.2(10)] -0.6[-1.4] -0.2[2 * (5-1)][ -[0.61]]$

$\frac{dT}{dt} = -0.8 +0.84 + 0.976$

$\frac{dT}{dt} = 1.016 ^oC/m$

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2 years ago

The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth. If a rock weighs 75.0 N on earth, compu

IgorLugansk [536]

Answer:

g = 0.905 gE

W = 67.9 N

Explanation:

given data

mass of Venus mv = 81.5% = 0.815

radius Rv = 94.9% = 0.949

weighs W = 75.0 N

solution

we apply here acceleration due to gravity at earth surface that is

g = $\frac{Gm}{R^2}$ = 9.80 m/s² ............1

so

g = $\frac{G(0.815)}{0.949R^2}$

g = 0.905 gE

and

W = m gv

W = 0.905 m gE

W = 0.905 × 75

W = 67.9 N

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2 years ago

A 50.0 kg crate is being pulled along a horizontal, smooth surface. The pulling force is 10.0 N and is directed 20.0 degree abov

allsm [11]

Explanation:

It is given that,

Mass of the crate, m = 50 kg

Force acting on the crate, F = 10 N

Angle with horizontal, $\theta=20^{\circ}$

Let N is the normal force acting on the crate. Using the free body diagram of the crate. It is clear that,

$N=mg-F\ sin\theta$

$N=50\times 9.8-10\ sin(20)$

N = 486.57 N

or

N = 487 N

If a is the acceleration of the crate. The horizontal component of force is balanced by the applied forces as :

$ma=F\ cos\theta$

$a=\dfrac{F\ cos\theta}{m}$

$a=\dfrac{10\times \ cos(20)}{50}$

$a=0.1879\ m/s^2$

or

$a=0.188\ m/s^2$

So, the normal force the crate and the magnitude of the acceleration of the crate is 487 N and $0.188\ m/s^2$ respectively.

4 0

2 years ago