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3 years ago

If Joey ran 60 meters in 10 seconds, then his velocity is ________ m/s.

Physics

2 answers:

BabaBlast [244]3 years ago

6 0

Answer:

6 m/s

Explanation:

BartSMP [9]3 years ago

4 0

The answer is : 6 m/s

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yulyashka [42]

Answer:

poda patiiiiiiiiii tadiya kundiii

5 0

2 years ago

A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia

Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0

3 years ago

One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched

mash [69]

Answer:

2.63 cm

Explanation:

Hooke's law gives that the force F is equal to cy where c is spring constant and x is extension

Making c the subject of the formula then

$c=\frac {F}{y}$

Since F is gm but taking the given mass to be F

$c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930$

By substitution now considering F to be 3.3 kg

$y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm$

8 0

3 years ago

A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with

gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm$

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s$