29 Family planning is a practice of regulating the number or spacing of offspring through the use of birth control.

A car, of mass 2.1 X 10^3 kg, travels in the horizontal plane around an unbanked curve of a radius of 275 m at a speed of 26 m/s

Nadusha1986 [10]

Answer:

Centripetal force is perpendicular to velocity and causes uniform circular motion. ... force exerted on a 900.0-kg car that negotiates a 500.0-m radius curve at 25.00 m/s. ... A car moving at 96.8 km/h travels around a circular curve of radius 182.9 m ... Because the car does not leave the surface of the road, the net vertical force ...

Explanation:

HE monda

4 0

2 years ago

A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i

vovangra [49]

Answer:

$c=71.4m/s$

$\theta=8.54\textdegree$

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed $V_w= 15 m/s \angle 45 \textdegree$

Generally Resolving vector mathematically

$sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6$

Generally the equation Pythagoras theorem is given mathematically by

$c^2=a^2+b^2$

$c^2=10.6^2 +(10.6+60)^2$

$c=\sqrt{10.6^2 +(10.6+60)^2}$

Therefore Resultant velocity (m/s)

$c=71.4m/s$

b)Resultant direction

Generally the equation for solving Resultant direction

$\theta=tan^-1(\frac{y}{x})$

Therefore

$\theta=tan^-1(\frac{10.6}{70.6})$

$\theta=8.54\textdegree$

7 0

2 years ago

Can someone tell me the answer to this?

Art [367]

Answer:

A

Explanation:

5 0

2 years ago

Read 2 more answers

A ship's wheel has a moment of inertia of 0.930 kilogram·meters squared. The inner radius of the ring is 26 centimeters, and the

Vikki [24]

We can use the formula of the moment of inertia given by:

$r\cdot F=I\alpha$

Where:

r = Distance from the point about which the torque is being measured to the point where the force is applied

F = Force

I = Moment of inertia

α = Angular acceleration

So:

$\begin{gathered} r\cdot F=(-0.26\times314+290\times0.32)=92.8-81.64=11.16 \\ I=0.930 \\ so,_{\text{ }}solve_{\text{ }}for_{\text{ }}\alpha: \\ \alpha=\frac{r\cdot F}{I} \\ \alpha=\frac{11.16}{0.930} \\ \alpha=\frac{12rad}{s^2} \end{gathered}$

Answer:

12 rad/s²

8 0

1 year ago

A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,