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viktelen [127]
3 years ago
11

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

m. "How long in seconds did it take for the box to slow by this amount, assuming the acceleration is constant?"
Physics
1 answer:
KiRa [710]3 years ago
8 0

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box=u=13m/s

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

v^2-u^2=2as

Substitute the values

(11.5)^2-(13)^2=2a(8.5)

132.25-169=17a

-36.75=17a

a=\frac{-36.75}{17}=-2.2m/s^2

We know that

Acceleration=a=\frac{v-u}{t}

Substitute the values

-2.2=\frac{11.5-13}{t}

-2.2=\frac{-1.5}{t}

t=\frac{1.5}{2.2}=0.68 s

Hence, the time taken by box to slow by this amount=0.68 s

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A compound microscope is made with an objective lens (f0 = 0.900 cm) and an eyepiece (fe = 1.10 cm). The lenses are separated by
Marizza181 [45]

Answer:

-252.52

Explanation:

L = Distance between lenses = 10 cm

D = Near point = 25 cm

f_o = Focal length of objective = 0.9 cm

f_e = Focal length of eyepiece = 1.1 cm

Magnification of a compound microscope is given by

m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52

The angular magnification of the compound microscope is -252.52

6 0
3 years ago
Starting from rest, a car undergoes a constant acceleration of 6 m/s^2. How far will the car travel in the two seconds?
oksano4ka [1.4K]

If time is specified, the distance may be estimated in constant acceleration using the formula: X=(at2)/2 if the beginning velocity is 0. (A automobile begins from a stop...) As a result, X=(6*10*10)/2=600/2 = 300 m.

4 0
2 years ago
A huge rotating cloud of particles in space gravitate together to form an increasingly dense ball. As ir shrinks in size, the cl
Trava [24]

Answer:

rotates faster

Explanation:

A huge rotating cloud of particles in space gravitate together to form an increasingly dense ball As it shrinks in size, the cloud rotates faster. Because Angular momentum is conserved, so when it shrinks the moment of inertia decreases, then angular speed must increase. So it rotates fast.

4 0
3 years ago
Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.
Juliette [100K]

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = \frac{mv}{qB}

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and  charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r =  \frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T} = 0.0328 m, or 3.28  cm.

8 0
3 years ago
A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th
vladimir1956 [14]

Answer:

Force, F=2.27\times 10^4\ N

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

F.d=\dfrac{1}{2}m(v^2-u^2)

Finally, it stops, v = 0

F.d=-\dfrac{1}{2}m(u^2)

F=\dfrac{-mu^2}{2d}

F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}

F = -22713.92 N

F=2.27\times 10^4\ N

So, the magnitude of the force that stops the bullet is 2.27\times 10^4\ N

7 0
3 years ago
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