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Tema [17]
3 years ago
12

Speed is the rate at which what happens?

Physics
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:

Speed is the rate at which an object covers distance

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Can y’all help me on number 2
Temka [501]

Answer:

sure

Explanation:

7 0
3 years ago
Read 2 more answers
A lizard accelerates from 2 m/s to 10 m/s in 4 seconds. what is the lizard average acceleration
VLD [36.1K]

Acceleration = (change in speed) / (time for the change)

change in speed = (ending speed) - (starting speed)

change in speed = (10 m/s) - (2 m/s)  =  8 m/s

Acceleration = (8 m/s) / (4 sec)

Acceleration = (8/4) (m/s²)

<em>Acceleration = 2 m/s²</em>

8 0
2 years ago
The momentum of a falling rock is found to be 200 kg m/s. What is the mass of the rock if it falls with a velocity of 5.0 m/s
Snezhnost [94]

Answer:

\boxed {\boxed {\sf 40 \ kilograms}}

Explanation:

Momentum is the product of velocity and mass. The formula is:

p=m*v

We know the rock is falling. Its momentum is 200 kilograms meters per second and its velocity is 5 meters per second. Substitute the values into the formula.

200 \ kg \ m/s = m * 5.0 \ m/s

We are solving for m, the mass. We must isolate the variable. It is being multiplied by 5 meters per second. The inverse of multiplication is division, so we divided both sides by 5.0 m/s.

\frac{200 \ kg \ m/s}{5.0 \ m/s}=\frac{ m* 5.0 \ m/s }{5.0 \ m/s}

\frac{200 \ kg \ m/s}{5.0 \ m/s}=m

The units of meters per second (m/s) cancel.

\frac{200 \ kg}{5.0 } =m

40 \ kg = m

The falling rock has a mass of <u>40 kilograms.</u>

4 0
2 years ago
I'LL GIVE BRAINLIEST TO THE FIRST OF BEST ANSWER
ira [324]
1.) because then people can evacuate the area in the path<span> of the hurricane.
2.) </span><span>At higher altitudes, water vapor starts to condense into clouds and rain, releasing heat that warms the surrounding air, Which makes it rise as well. Warmer waters feed more energetic storms.
3.) </span> <span>A hurricane starts off as a series of thunderstorms which intensify as it moves over the warm and humid sea. The humidity is at a constant level and so it continues to grow over the sea. Any kind of decrease or increase in humidity can change the strength of a hurricane. 
4.) </span><span>Actually, tropical cyclones need weak winds. If the atmospheric winds are even remotely strong, they will act to cut back the system and prevent the convection from wrapping around the center. 
</span><span>Annndd... 
5.) That hard to tell, it could be too much. Though I am going to go with yes. Cyclones need weak winds and good amount humidity.</span><span>




</span>
5 0
3 years ago
A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region
zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with  due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

K_{1} = K_{2} + W_{f}

Where:

K_{1}, K_{2} are the initial and final translational kinetic energies of the tobbogan, measured in joules.

W_{f} - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})

Where:

f - Friction force, measured in newtons.

\Delta s - Distance travelled by the toboggan in the rough region, measured in meters.

m - Mass of the toboggan, measured in kilograms.

v_{1}, v_{2} - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}

If m = 375\,kg, v_{1} = 4.50\,\frac{m}{s}, v_{2} = 1.20\,\frac{m}{s} and \Delta s = 5.40 \,m, then:

f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}

f = 653.125\,N

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%

\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%

\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%

\%K_{loss} = 92.889\,\%

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%

\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%

\%v_{loss} = 73.333\,\%

The speed of the toboggan is reduced in 73.333 %.

5 0
3 years ago
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