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3 years ago

14

A stack of books rests on a level frictionless surface. A force F acts on the stack, and it accelerates at 3.0 m/s2. A 1.0 kg bo

ok is then added to the stack. The same force is applied, and now the stack accelerates at 2.0 m/s2.

1 answer:

Oxana [17]3 years ago

4 0

Answer:

m1 = 2 kg

m2 = 3kg

Explanation:

The force can be getting by

F = m * a

F1 = F2

a1 = 3.0 m/s^2

a2 = 2.0 m/s^2

The force F1=F2 because the force is applied so get the a2 acceleration

m1 * a1 = m2 * a2

m2 = m1 + 1kg

m1 *(3.0 m/s^2) = m2* (2.0 m/s^2)

m1 *(3.0 m/s^2) = (m1 + 1kg) * (2.0 m/s^2)

m1*(3.0m/s^2-2.0m/s^2)=2 kg*m/s^2

Solve to find the mass

m1 m/s^2= 2 kg*m/s^2

m1 = 2 kg

m2 = 3kg

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MA_775_DIABLO [31]

Answer:

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2 years ago

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

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1 year ago

If the wavelength of light in the visible region is known, what is also known? check all that apply. question 4 options:

maw [93]

If the wavelength of light in the visible region is known, it is also known as frequency.

<h3>What is frequency?</h3>

Recurrence is the quantity of events of a rehashing occasion for every unit of time. It is likewise once in a while alluded to as worldly recurrence to underline the differentiation to spatial recurrence, and customary recurrence to underscore the difference to rakish recurrence. Rotating current (ac) recurrence is the quantity of cycles each second in an air conditioner sine wave. Recurrence is the rate at which current heads in a different path each second. It is estimated in hertz (Hz), a global unit of measure where 1 hertz is equivalent to 1 cycle each second. It is likewise infrequently alluded to as transient recurrence to stress the difference to spatial recurrence, and normal recurrence to accentuate the differentiation to precise recurrence. Recurrence is estimated in hertz (Hz) which is equivalent to one occasion each second.

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1 year ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

2 years ago

Jessica stretches her arms out 0.60 m from the center of her body while holding a 2.0 kg mass in each hand. She then spins aroun

Juliette [100K]

Answer:

a.) L = 2.64 kgm^2/s

b.) V = 4.4 m/s

Explanation: Jessica stretches her arms out 0.60 m from the center of her body. This will be considered as radius.

So,

Radius r = 0.6 m

Mass M = 2 kg

Velocity V = 1.1 m/s

Angular momentum L can be expressed as;

L = MVr

Substitute all the parameters into the formula

L = 2 × 1.1 × 0.6 = 1.32kgm^2s^-1

the combined angular momentum of the masses will be 2 × 1.32 = 2.64 kgm^2s-1

b. If she pulls her arms into 0.15 m,

New radius = 0.15 m

Using the same formula again

L = 2( MVr)

2.64 = 2( 2 × V × 0.15 )

1.32 = 0.3 V

V = 1.32/0.3

V = 4.4 m/s

Her new linear speed will be 4.4 m/s

4 0

2 years ago