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Oduvanchick [21]
3 years ago
10

a cruise took 6 hours to fulfill its trip it travelled at an average speed of 160 km/h for thefirst 2 h 15 min

Physics
1 answer:
soldi70 [24.7K]3 years ago
6 0

Answer: 1035

Explanation:

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A Truck with a mass of 1500 kg is decelerated At a rate of 5m/s2. how much force did this require
Marianna [84]
(1500 kg)*(5 m/s^2) = 7500 N
6 0
3 years ago
B
SVEN [57.7K]

Answer:

Frequency = 1,550Hz

Explanation:

To solve this we can use the equation: f=\frac{v}{\lambda}

(frequency = velocity/wavelength).

We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m

Now we can substitute these values into the formula and calculate to solve:

f=\frac{340}{0.22} \\\\f=1545.454...

Simplify to 3 significant figures:

f = 1,550Hz

(Which I believe is just below a G6 if you were interested)

Hope this helped!

4 0
3 years ago
Light striking a mirror at a 50° angle will be reflected at an angle _____.
Stells [14]
Equal to 50


law of reflection: angle of incidence equals angle of reflection
5 0
3 years ago
Read 2 more answers
Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur
Alenkasestr [34]

Answer:

the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}

the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder  be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be \phi and \beta respectively.

The velocity of the block to be = v

The equivalent mass of the system = m_e

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

K.E = \frac{1}{2} m_ev^2       --------------- equation (1)

The angular velocity of the cylinder = \omega  :  &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

v = \omega R

\omega =\frac{v}{R}

The kinetic energy of the system in terms of individual mass can be written as:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2

By replacing \omega with \frac{v}{R} ; we have:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2

K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2   ------------------ equation (2)

Equating both equation (1) and (2); we have:

m_e = m_1+m_2+\frac{I}{R^2}

Therefore, the equivalent mass : m_e = m_1+m_2+\frac{I}{R^2}    which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by  the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

f_{block }=m_1gsin \beta

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

f_{cylinder} = m_2gsin \phi

Equating the above both equations; we have the equation of motion of the  equivalent system to be

m_e \bar x = f_{cylinder}-f_{block}

which can be written as follows from the previous derivations

(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Finally; the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

8 0
3 years ago
The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:
Furkat [3]
To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is  s=r(B), that has its <span>r=radius,B=angle in rad velocity v=ds/dt= w(r)
Do not forget about </span> w = angular speed in rad/s and w1 = 1 revolution/sec = 2Pi (rad/s)
Now we can go to proportion
v1=v2
w1*r1 = w2r2w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)
w2 = w3 (which is the   angular velocity of the rear wheel) &#10;
SOLVING FOR A : v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec
v3 = 14.66 ft/sec(1 mile/5280 ft)( 3600 sec/h)= 9.99 or something about <span>10 mph --- SOLVING FOR B.
</span>I'm sure it helps!
7 0
3 years ago
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