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3 years ago

14

A 2000kg suv accelerates from rest at a rate of 3.00m/s^2. The total amount of force resisting its motion 1500N. How much force

is applied to the suvs tires by the ground to make it accelerate?

1 answer:

choli [55]3 years ago

7 0

The total force that the SUV exerts is:

F = 2000 kg * 3 m/s^2

F = 6000 N

Since a resisting force amounting to 1500 N is exerted, then the force exerted by the SUV tires is:

F tire = 6000 N – 1500 N

F tire = 4500 N

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Why were epicycles necessary in ptolemy’s model of the universe?

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Calculate the speed for a car that went a distance of 125 meters in 2 seconds time.

stiv31 [10]

Answer:

<h2>62.5 m/s</h2>

Explanation:

The speed of the car can be found by using the formula

$s = \frac{d}{t} \\$

d is the distance

t is the time

From the question we have

$s = \frac{125}{2} = 62.5 \\$

We have the final answer as

<h3>62.5 m/s</h3>

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2 years ago

The universal law of gravitation states that the force of attraction between two objects depends on which quantities?

Ronch [10]

Answer:

D. the masses of the objects and the distance between them

Explanation:

Gravitation is a force, a force doesn't care about the shape or density of objects, only about their masses... and distances.

And you can get it using the following equation:

$f = \frac{Gm_{1}m_{2} }{d^{2} }$

Where :

G is the universal gravitational constant : G = 6.6726 x 10-11N-m2/kg2

m represent the mass of each of the two objects

d is the distance between the centers of the objects.

4 0

3 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

3 years ago

I need help ASAP. This is for 15 points

kramer

Answer:

Latitude :

runs: east to west

measures : distances north and south of the equator

Longitude :

runs : north to south

measures : the distance east or west of the Prime Meridian

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3 years ago