As a ball bounces from a floor, its acceleration off the floor between bounces is

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th

jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

R = Vo² sin (2θ) / g

sin 2θ = g R / Vo²

sin 2θ = 9.8 75/35²

2θ = sin⁻¹ (0.6)

θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

X = Vox t

t2 = X2 / Vox = X2 / (Vo cosθ)

t2 = 37.5 / (35 cos 18.4)

t2 = 1.13 s

With this time we calculate the height at this point

Y = Voy t - ½ g t²

Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²

Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

8 0

2 years ago

Read 2 more answers

A certain car battery with a 12.0 V emf has an initial charge of 131 A · h. Assuming that the potential across the terminals sta

irga5000 [103]

Answer:

The battery can supply 130 W for 11.75 h

Explanation:

In order to discover the time in wich the battery can supply this energy we need to find how much current is being drawn from it, we do that by using the equation for real power that is P = V*I, since we have V and P we can solve for I as seen bellow:

I = P/V = 130/12 = 10.834 A

We can use this value to find how many hours the power can supply said current. We do that by dividing the current capacity of the battery by the current drawn:

t = 141/12 = 11.75 h

7 0

3 years ago

PLEASE HELP! It’s urgent... and please show your work!!

bagirrra123 [75]

Answer:

A) 3.8 x 10

Explanation:

4 0

2 years ago

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

$E=\dfrac{\sigma}{2\epsilon_0}$

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Now, plug in $47\times 10^{-9}\ C/m^2$ for 'σ' and $8.85\times 10^{-12}\ F/m$ for $\epsilon_0$ and solve for the electric field. This gives,

$E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C$

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0

2 years ago

Draw additional masses to show how the block can be made to move toward point A. Be sure to draw the string and label the mass o

jeyben [28]

With 250 grams pullling to the right and 100 grams pulling towards A, you will have some movement towards the right and some movement towards A. The resultant vector will be angled between those two directions.

5 0

3 years ago