HURRY ILL GIVE 15 POINTS

Which of the following are good safety measures during a thunderstorm? A. Talk on a phone with a cord. B. Go inside a closed bui

Alex73 [517]

B and C are the correct answers because when you stand in an open space the ground is an conductor therefore adding and electric shock to your day

7 0

3 years ago

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The radius of he Earth orbit around the sun (assumed circular) is 1.50 X 10^8km, with T=365d. What is the radial acceleration of

Karolina [17]

Answer:

ar = 5.86*10^-3 m/s^2

Explanation:

In order to calculate the radial acceleration of the Earth, you first take into account the linear speed of the Earth in its orbit.

You use the following formula:

$v=\sqrt{\frac{GM_s}{r}}$ (1)

G: Cavendish's constant = 6.67*10^-11 m^3 kg^-1 s^-2

Ms: Sun's mass = 1.98*10^30 kg

r: distance between Sun ad Earth = 1.50*10^8 km = 1.50*10^11 m

Furthermore, you take into account that the radial acceleration is given by:

$a_r=\frac{v^2}{r}$ (2)

You replace the equation (1) into the equation (2) and replace the values of all parameters:

$a_r=\frac{1}{r}\frac{GM_s}{r}=\frac{GM_s}{r^2}\\\\a_r=\frac{(6.67*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}{(1.50*10^{11}m)^2}\\\\a_r=5.86*10^{-3}\frac{m}{s^2}$

The radial acceleration of the Earth, towards the sun is 5.86*10^-3 m/s^2

7 0

3 years ago

What are the horizontal and vertical velocities of a stunt bike that leaves a ramp at 100 km/hr and at an angle of 35 degrees?

Alex787 [66]

Horizontal velocity: 81.9 km/h

Vertical velocity: 57.4 km/h

Explanation:

We can solve this problem by resolving the velocity vector into its component along the horizontal and vertical direction.

The horizontal velocity of the stunt bike is given by:

$v_x = v cos \theta$

where

v = 100 km/h is the magnitude of the velocity

$\theta=35^{\circ}$ is the angle of projection

Substituting, we find

$v_x = (100)(cos 35^{\circ})=81.9 km/h$

The vertical velocity instead is given by

$v_y = v sin \theta$

where

$v=100 km/h$

$\theta=35^{\circ}$

Substituting,

$v_y = (100)(sin 35^{\circ})=57.4 km/h$

Learn more about vector components:

brainly.com/question/2678571

#LearnwithBrainly

4 0

3 years ago

A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of

Sunny_sXe [5.5K]

Answer:

$5.8\; {\rm kg\cdot m \cdot s^{-1}}$ .

Explanation:

If the mass of an object is $m$ and the velocity of that object is $v$ , the linear momentum of that object would be $m\, v$ .

Assume that the initial velocity of the mass is positive ( $6.0\; {\rm m\cdot s^{-1}}$ .) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be $(-4.0\; {\rm m\cdot s^{-1}})$ .

Thus, the change in the velocity of the mass would be:

$\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}$ .

The change in the linear momentum of the mass would be:

$\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\ =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}$ .

Thus, the magnitude of the change of the linear momentum would be $5.8\; {\rm kg \cdot m \cdot s^{-1}}$ .

7 0

3 years ago

The prefrontal cortex helps in

Lisa [10]

A i think its A I think

4 0

4 years ago