A 2000 kg car is traveling at 10 m/s, and then it slows to 5 m/s. How much work is done on the car by the brakes and friction?

You've recently read about a chemical laser that generates a 20.0-cm-diameter, 26.0 MW laser beam. One day, after physics class,

aksik [14]

Answer :

(a). The speed of the block is 0.395 m/s.

(b). No

Explanation :

Given that,

Diameter = 20.0 cm

Power = 26.0 MW

Mass = 110 kg

diameter = 20.0 cm

Distance = 100 m

We need to calculate the pressure due to laser

Using formula of pressure

$P_{r}=\dfrac{I}{c}$

$P_{r}=\dfrac{P}{Ac}Put the value into the formula[tex]P_{r}=\dfrac{26.0\times10^{6}}{\pi\times(10\times10^{-2})^2\times3\times10^{8}}$

$P_{r}=2.75\ N/m^2$

We need to calculate the force

Using formula of force

$F=P\times A$

$F=P\times \pi r^2$

Put the value into the formula

$F=2.75\times\pi (0.01)^2$

$F=0.086\ N$

We need to calculate the acceleration

Using formula of force

$F=ma$

Put the value into the formula

$0.086=110\times a$

$a=\dfrac{0.086}{110}$

$a=0.000781\ m/s^2$

$a=7.81\times10^{-4}\ m/s^2$

(a). We need to calculate speed of the block

Using equation of motion

$v^2=u^2+2ad$

Put the value into the formula

$v=\sqrt{2\times7.81\times10^{-4}\times100}$

$v=0.395\ m/s$

(b). No because the velocity is very less.

Hence, (a). The speed of the block is 0.395 m/s.

(b). No

8 0

3 years ago

Give examples of plants that can reproduce by vegetative reproduction.

marishachu [46]

Answer:

Some examples of vegetative propagation are farmers creating repeated crops of apples, corn, mangoes or avocados through asexual plant reproduction rather than planting seeds. Vegetative propagation can be accomplished from side-shoots, slips, stems and sections of tubers, bulbs or rhizomes.

Explanation:

5 0

4 years ago

Three heavy rods are all made of the same uniform material. These rods have lengths 3 m, 4 m, and 5 m, and compose the sides of

Sonbull [250]

Answer:

[ 2.67 , 1 ] m

Explanation:

Given:-

- The side lengths of the rods are as follows:

a = 4 m , b = 4 m , c = 5 m

a = Base , b = Perpendicular , c = Hypotenuse

- All rods are made of same material with uniform density. With

Find:-

Find the coordinates of the center of mass of the triangle.

Solution:-

- The center of mass of any triangle is at the intersection of its medians.

- So let’s say we have a triangle with vertices at points (0,0) , (a,0) , and (0,b).

Median from (0,0) to midpoint (a/2,b/2) of opposite side has equation:

bx−ay=0

Median from (a,0) to midpoint (0,b/2) of opposite side has equation:

bx+2ay=ab

Median from (0,b) to midpoint (a/2,0) of opposite side has equation:

2bx+ay=ab

Solve all three equations simultaneously:

bx−ay=0 , bx = ay

ay + 2ay = ab , 3ay = ab , y = b/3

bx = b/3

x = a / 3

So the distance from the median to each leg of the triangle is 1/3 length of other leg.

- So the coordinates of the centroid for right angle triangle would be:

[ 2a/3 , b/3 ]

[ 2.67 , 1 ] m

3 0

3 years ago

A transport truck pulls on a trailer with a force of 600N [E]. The trailer pulls on the transport truck with a force

kramer

These forces form a force pair. Use Newton's third law, and you see that the trailer pulls back at with the same force. The answer is d.

6 0

3 years ago

Read 2 more answers

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
 $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
 $E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
 $E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
 $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
 $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
 $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
 $E=U+K$
we find the kinetic energy when the glider is at this position:
 $K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
 $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
 $a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
 $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

we have already calculated it at point b), and it is given by
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

4 years ago