What is the distance a bullet will travel when launched at 166 meters per second for 5.75seconds?

Both positive and negative charges will spread the leaves of an uncharged electroscope.

hoa [83]

The answer is A)TRUE.

7 0

3 years ago

If the distance between the two objects is reduced in half, what will be the changed force of attraction between them?

NNADVOKAT [17]

Will reduce 4 times.

$F = G \frac{m1* m2}{ d^{2} }$
So called LUG (Law of Universal gravitation).

8 0

3 years ago

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo

Jobisdone [24]

(a) 4.0 m/s

We can solve this part just by analyzing the vertical motion of the froghopper.

The initial vertical velocity of the froghopper as it jumps from the ground is given by

$u_y = u_0 sin \theta$ (1)

where

$u_0$ is the takeoff speed

$\theta=58.0^{\circ}$ is the angle of takeoff

The maximum height reached by the froghopper is

h = 58.7 cm = 0.587 m

We know that at the point of maximum height, the vertical velocity is zero:

$v_y = 0$

Since the vertical motion is an accelerated motion with constant (de)celeration $g=-9.8 m/s^2$ , we can use the following SUVAT equation:

$v_y^2 - u_y^2 = 2gh$

Solving for $u_y$ ,

$u_y = \sqrt{v_y^2-2gh}=\sqrt{-2(-9.8)(0.587)}=3.4 m/s$

And using eq.(1), we can now find the initial takeoff speed:

$u_0 = \frac{u_y}{sin \theta}=\frac{3.4}{sin 58.0^{\circ}}=4.0 m/s$

(b) 1.47 m

For this part, we have to analyze the horizontal motion of the froghopper.

The horizontal velocity of the froghopper is

$u_x = u_0 cos \theta = (4.0) cos 58.0^{\circ} =2.1 m/s$

And this horizontal velocity is constant during the entire motion.

We now have to calculate the time the froghopper takes to reach the ground: this is equal to twice the time it takes to reach the maximum height.

The time needed to reach the maximum height can be found through the equation

$v_y = u_y + gt$

Solving for t,

$t=-\frac{u_y}{g}=-\frac{3.4}{9.8}=0.35 s$

So the time the froghopper takes to reach the ground is

$T=2t=2(0.35)=0.70 s$

And since the horizontal motion is a uniform motion, we can now find the horizontal distance covered:

$d=u_x T = (2.1)(0.70)=1.47 m$

7 0

3 years ago

). with the input voltage range set at +/- 500mv, what is the smallest difference in voltage that can be resolved? show your cal

dalvyx [7]

The smallest difference in voltage that can be resolved is referred to as the resolution. The resolution can be calculated with the following formula:
resolution=voltage range / digital range
The voltage range in our case is from -500mV to 500mV, which gives 1000mV.
The digital range on the other hand is 2^(number of bits).
It depends on what type of bit board we are using. If the ADC we are using is a 16 bit board, then 2^16=<span>65536.
So, the resolution is:
resolution=1000mV/</span><span>65536=0.015 mV</span>

4 0

3 years ago

The maximum amount of pulling force a truck can apply when driving on

kupik [55]

C the correct but not sure?

5 0

3 years ago