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3 years ago

The race car was was moving for 3.7 hours and during that time moved 500 miles east. What is the speed? What is the velocity?

Physics

1 answer:

vaieri [72.5K]3 years ago

8 0

Speed:

$v = \frac{500 \,\,\mbox{miles}}{3.7 \,\,\mbox{h}}\approx 135.14 mph$

Velocity:

Velocity os speed with director (an oriented vector):

The velocity is 135.14 mph in the direction east

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A 1,000 kg car is driving on a 15 m high bridge at 5 m/s. What is the kinetic energy of the car?

yanalaym [24]

Answer:

$KE=12,500J$

Explanation:

The formula for kinetic energy is:

$KE = \frac{1}{2}mv^2$

We can plug in the given values into the equation:

$KE = \frac{1}{2}*1000kg*(5m/s)^2$

$KE = 500kg*25m^2/s^2$

$KE=12,500J$

7 0

2 years ago

A fan cart initially has an acceleration of 1.6m/s/s when it's fan is directed straight backwards. If you rotate the fan by 45°,

Sholpan [36]

Answer:

A Fan Cart Initially Has An Acceleration Of 1.6m/s/s When It's Fan Is Directed Straight Backwards. If You Rotate The Fan By 45o, By What Percentage Do You Expect The Fan Cart's Thrust To Decrease? (Answer Should Be In Units Of 96)

a. 45%

b. 29%

c. 71%

d. 50%

The correct answer is d.

d. 50%

Explanation:

Fan cart acceleration = 1.6 m/s²

Thrust = 0.25×π×D²×ρ×v×Δv

where Δv = acceleration component and all factors remaining cconstant, when the fan is rotated by 45 ° the diameter changes to D₂ = sin 45 ×D

or 0.707×D. The thrust becomes 0.25×π×(0.707×D)²×ρ×v×Δv

=0.25×π×0.5×D²×ρ×v×Δv or 0.5(0.25×π×D²×ρ×v×Δv)

That is the thrust reduces by 50 %

3 0

3 years ago

Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M

stealth61 [152]

The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

We have to write the equations of motion for the two blocks.

- For M1, the only two forces acting on it are the force of gravity $M_1 g$ (downward) and the tension in the string T (upward). So we can write

$M_1 g - T = M_1 a$

where

$M_1$ is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$a$ is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, $M_2 g sin \theta$ . So the equation of motion is

$T-M_2 g sin \theta = M_2 a$

where

$M_2 = 13.5 kg$ is the mass of the 2nd block

$\theta=35.5^{\circ}$ is the angle of the ramp

In order for the two blocks to be in equilibrium, the acceleration must be zero:

$a=0$

So the two equations become:

$M_1 g - T=0\\T-M_2 g sin \theta = 0$

Isolating T from the 1st equation,

$T=M_1 g$

And substituting into the 2nd equation, we can find the value of the mass $M_1$ :

$M_1 g - M_2 g sin \theta = 0\\M_1 = M_2 sin \theta = (13.5)(sin 35.5^{\circ})=7.8 kg$

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

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#LearnwithBrainly

7 0

3 years ago

A difference In density between gases creates buoyancy which is the ability of an object to float. 0°C helium has a density of .

rjkz [21]

Answer:

mass = 0.18 [kg]

Explanation:

This is a classic problem where we can apply the definition of density which is equal to mass over volume.

$density = \frac{mass}{volume} \\\\where:\\volume = 1 [m^3]\\density = 0.18[kg/m^3]$

mass = 0.18*1

mass = 0.18 [kg]

6 0

3 years ago

F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$