On earth which force is 10 to slow an object down

The mass of a sports car is 1000 kg. The shape of the car is such that the aerodynamic drag coefficient is 0.260 and the frontal

alisha [4.7K]

Answer:

The initial acceleration is 0,221 m/s^2.

Explanation:

$Fcar = Fairfriction\\m*a = 0.5*A*c*density*v^2\\a = 0.5*A*c*density*v^2 / 1000 kg = 0.5*2.10m^2*0.260*1.295kg/m^3*(25 m/s)^2 / 1000 kg = 0,221 m/s^2$

8 0

3 years ago

Read 2 more answers

A car of mass M = 1000 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ , an

kupik [55]

The radius of the curved road at the given condition is 54.1 m.

The given parameters:

mass of the car, m = 1000 kg
speed of the car, v = 50 km/h = 13.89 m/s
banking angle, θ = 20⁰

The normal force on the car due to banking curve is calculated as follows;

$Fcos(\theta) = mg$

The horizontal force on the car due to the banking curve is calculated as follows;

$Fsin(\theta) = \frac{mv^2}{r}$

Divide the second equation by the first;

$\frac{Fsin(\theta)}{Fcos(\theta) } = \frac{mv^2}{rmg} \\\\tan(\theta) = \frac{v^2}{rg} \\\\r = \frac{v^2}{g \times tan(\theta)} \\\\r = \frac{13.89^2}{9.8 \times tan(20)} \\\\r = 54.1 \ m$

Thus, the radius of the curved road at the given condition is 54.1 m.

Learn more about banking angle here: brainly.com/question/8169892

3 0

3 years ago

A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag

Readme [11.4K]

Answer:

(a)0.0675 J

(b)0.0675 J

(c)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

$E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675 J$

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

4 0

3 years ago

Hanging from a horizontal beam are nine simple pendulums of the following lengths:

LenaWriter [7]

Answer:

Options d and e

Explanation:

The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.

We can get the length of the pendulums likely to oscillate with the formula;

$L =\frac{g}{w^{2} }$

where g=9.8m/s

ω= 2rad/s to 4rad/sec

when ω= 2rad/sec

$L= \frac{9.8}{2^{2} }$

L = 2.45m

when ω= 4rad/sec

$L=\frac{9.8}{4^{2} }$

L = 9.8/16

L=0.6125m

L is between 0.6125m and 2.45m.

This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.

Have a great day ahead

8 0

4 years ago

Caregivers who are observed to be abusive or neglectful have been associated with the __________ type of attachment. A. secure B

Whitepunk [10]

Crrfhhvhhhhhjvvcffggh

4 0

3 years ago