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ella [17]
3 years ago
11

A book is sitting on a desk. What best describes the normal force acting on the book?​

Physics
2 answers:
maxonik [38]3 years ago
8 0

Gravitational force is the result of the earth pulling down on the book, so the normal force is best described as the earth or more accurately the table pushing up on the book

Alex17521 [72]3 years ago
3 0

Answer:

The downward force the Earth pulls on the book

Explanation:

If the force was pulling any other direction the book would not be resting on the table.

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The linear expansivity of metal P is twice that of another metal Q. When these materials are heated through the same temperature
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3 years ago
A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has coordinates and . What is the
Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

$y = x^2+4x+6m$

$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$. What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1 $

$x=2 $

and $\rho = 2 \ kg/m^2$ , $P_x=1 \ $ , $P_y=-2 \ $.

So,

$dI = dmr^2$

$dI = \rho \ dA  \ r^2$  ,           $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \  dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

  $= 3050.19 \ kg \ m^2$

So the moment of inertia is  $3050.19 \ kg \ m^2$.

4 0
3 years ago
Which term refers to a variable that a scientist adjusts during an experiment?
azamat

Answer:

it’s c

Explanation:

5 0
3 years ago
A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.08190.0819 m, its frequency
Anvisha [2.4K]

Answer:

The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

Explanation:

Given that,

Amplitude = 0.08190 m

Frequency = 2.29 Hz

Wavelength = 1.87 m

(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave

Using formula of distance

d=2A

Where, d = distance

A = amplitude

Put the value into the formula

d=2\times0.08190

d=0.1638\ m

Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

5 0
3 years ago
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