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3 years ago

A 0.75 kg book is pushed across the table with an acceleration of 0.3 m/s2. What force is being applied to the

Physics

1 answer:

Brums [2.3K]3 years ago

5 0

Answer:

$\boxed {\boxed {\sf 0.225 \ Newtons}}$

Explanation:

We are asked to find the force being applied to a book. According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F= ma$

The mass of the book is 0.75 kilograms and the acceleration is 0.3 meters per square second. Substitute these values into the formula.

m= 0.75 kg
a= 0.3 m/s²

$F= 0.75 \ kg * 0.3 \ m/s^2$

Multiply.

$F =0.225 \ kg * m/s^2$

1 kilogram meter per second squared is equal to 1 Newton. Therefore, our answer of 0.225 kilogram meters per second squared is equal to 0.225 Newtons.

$F= 0.225 \ N$

0.225 Newtons of force are applied to the book.

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Bradley gets an x-ray at a radiology clinic that employs its own technologists and radiologists. Would the coder at the clinic r

KengaRu [80]

Answer:

Explanation:

If Bradley examination was done and interpreted in the same facility, the radiologist code is used example- procedure code 72100- Radiologic examination, spine, lumbosacral, 2 or 3 views is reported.

if the X-ray was taken by Dr X but Dr X does not read or interpret the image but forward it to the radiologist for initial report, then a 26- modifier is used. E.g A reports by the technologist would be, procedure code 72050-Radiologic examination, spine, cervical, 2 or 3

views or 72050- TC in certain situations and the consulting radiologist would report 72050-26.

if Bradley’s x-ray were sent to an independent radiologist for interpretation, then the procedure code 76140 is used in reporting.

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2 years ago

A4 40 kg girl skates at 3.5 m/s one ice toward her 65 kg friend who is standing still, with open arms. As they collide and hold

salantis [7]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

Lorentz transformation

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago

Sound waves travel fastest through a A) gas. B) liquid. C) solid. D) vacuum.

earnstyle [38]

Sound waves travel faster through solids than they do through gases or liquids. (C) They don't travel through vacuum at all.

Example:

Speed of sound in normal air . . . around 340 m/s

Speed of sound in water . . . around 1,480 m/s

Speed of sound in iron . . . around 5,120 m/s

3 0

3 years ago

Read 2 more answers

The angular width of localizers varies between ______ and ______ degrees in order to provide a signal width of approximately 700

natali 33 [55]

Answer: 10 and 35 degrees

Explanation: Localizers width below 10 degree and 35 degree signal arc is unreliable and considered unusable for navigation and as a result, aircrafts may loose alignment

5 0

3 years ago