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blondinia [14]
3 years ago
13

A Formula 1 racing car can accelerate from a stationary position to 25 m/S in 2.1s calculate

Physics
1 answer:
zlopas [31]3 years ago
5 0

Answer:

120 m/s²

Explanation:

25 – 0 / 2.1 = 119.0476 = 120 m/s² to two significant figures

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Football player A has a mass of 210 pounds and is running at a rate of 5.0 mi/hr. He collides with player B. Player B has a mass
Naddika [18.5K]

The equation for momentum is p = mv where p is the omentum, m is the mass and v is the velocity. Calculating the momentum for each football player, player A will have a momentum of 1050 lb-mi/h and player B will have a momentum of 570 lb-mi/h. Therefore, momentum of player A is greater than that of player B.

6 0
3 years ago
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What voltage would produce a current of 100amps through an aluminum wire assuming that the re sistance of the wire is3.44×10-4 o
agasfer [191]
V = IR

I = current
R = resistance

Voltage = 100 * (3.44x 10^-4) = do the calculation

Hope this helps
8 0
3 years ago
A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0
3 years ago
Uest<br>1. State Newton's law of cooling.​
garik1379 [7]

Answer:

Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. The law is frequently qualified to include the condition that the temperature difference is small and the nature of heat transfer mechanism remains the same. As such, it is equivalent to a statement that the heat transfer coefficient, which mediates between heat losses and temperature differences, is a constant. This condition is generally met in heat conduction (where it is guaranteed by Fourier's law) as the thermal conductivity of most materials is only weakly dependent on temperature. In convective heat transfer, Newton's Law is followed for forced air or pumped fluid cooling, where the properties of the fluid do not vary strongly with temperature, but it is only approximately true for buoyancy-driven convection, where the velocity of the flow increases with temperature difference. Finally, in the case of heat transfer by thermal radiation, Newton's law of cooling holds only for very small temperature differences.

When stated in terms of temperature differences, Newton's law (with several further simplifying assumptions, such as a low Biot number and a temperature-independent heat capacity) results in a simple differential equation expressing temperature-difference as a function of time. The solution to that equation describes an exponential decrease of temperature-difference over time. This characteristic decay of the temperature-difference is also associated with Newton's law of cooling

6 0
3 years ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
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