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3 years ago

9

How do you calculate speed

2 answers:

Kryger [21]3 years ago

8 0

Answer:

distance(d)=.....

time(t)=....

speed =?

we know that

speed =distance /time

Amanda [17]3 years ago

5 0

The formula for speed is speed = distance ÷ time.

Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (Δt), represented by the equation r = d/Δt.

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A mass of 2000 kg. is raised 2.0 m in 10 seconds. What is the kinetic energy of the mass at this height?

Stella [2.4K]

The K.E. of that mass at this height is 40J.

7 0

3 years ago

Calculate the density of an object with a volume of 18m3 and a mass of 3.5kg?

jasenka [17]

0.19
Explanation:you will divide 3.5 and 18

3 0

3 years ago

A cart of mass 6.0 kg moves with a speed of 3.0 m/s towards a second stationary cart with a mass of 3.0 kg. The carts move on a

katen-ka-za [31]

Answer:b

Explanation:

Given

mass of first cart $m_1=6 kg$

mass of second cart $m_2=3 kg$

velocity of first cart $v_1=3 m/s$

conserving momentum

$m_1v_1+m_2v_2=(m_1+m_2)v$

$6\times 3+3\times 0=(9)\cdot v$

$v=2 m/s$

Initial kinetic Energy $K.E._1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$

$K.E._1=\frac{1}{2}\cdot 6\cdot 3^2+0$

$K.E._1=27 J$

Final Kinetic Energy

$K.E._2=\frac{1}{2}(m_1+m_2)v^2$

$K.E._2=\frac{1}{2}(6+3)\cdot 2^2=18 J$

Ratio of initial Kinetic Energy to the Final Kinetic Energy

$=\frac{27}{18}=1.5$

6 0

4 years ago

A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli

fredd [130]

Answer:

Part a)

$A = 1.78 m$

Part b)

$f = 2 rev/s$

Explanation:

Part A)

As we know that time period of the motion is given as

$T = 2.68 s$

so we have

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.68}$

$\omega = 2.34 rad/s$

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

$mg = m\omega^2 A$

so we have

$A = \frac{g}{\omega^2}$

$A = \frac{9.81}{2.34^2}$

$A = 1.78 m$

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

$mg = m\omega^2 A$

$g = \omega^2 (0.0623)$

$\omega = 12.5 rad/s$

so now we have

$2\pi f = 12.5$

$f = 2 rev/s$

3 0

3 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago