Gianna walked at 22m/s and travelled 110 meters. How long did it take Gianna to complete her trip?

A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

4 years ago

A geosynchronous satellite orbits above the equator of Earth in a circular orbit, remaining in a fixed position with respect to

Anestetic [448]

Answer:

Explanation:

The relation between orbital period T and orbital radius R is as follows .

T² ∝ R³

T ∝ R¹°⁵

So time period of orbit is proportional to radius of orbit . Higher the height , larger the orbital period . As the orbital period is larger than required , the altitude of satellite must have been larger than required .

As mass of satellite is not involved in the formula of orbital period , this is independent of mass of the satellite .

Hence the option C is correct .

8 0

3 years ago

A 217 kg motorcycle moves with a velocity of 2 m/s. what is it's kinetic energy?

SpyIntel [72]

The answer is 434J because kinetic energy is1/2Mv2

5 0

3 years ago

Using the superposition method, calculate the current through R5 in Figure 8-71

Natasha2012 [34]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

$r_1 = \frac{2.2 * 1}{2.2 + 1}$

$r_1 = 0.6875 k ohm$

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

$r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}$

$r_2 = 0.95 k ohm$

now the current flowing through the battery is

$i = \frac{2}{1 + 0.95} = 1.02 mA$

now this will divide into R3 and R2 so current flowing in R3 will be

$i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA$

now this will again divide in R4 and R5

so current in R5 will be

$i_5 = \frac{R_4}{R_4 + R_5}* i_1$

$i_5 = 0.18 mA$

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

$r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm$

also after its series with R3

$r_2 = 1 + 0.6875 = 1.6875 k ohm$

now it is in parallel with R5 on other side

$r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm$

now current through the battery will be given as

$i = \frac{3}{1 + 0.95} = 1.53 mA$

now it is divide in r2 and R5

so current in R5 is given as

$i_5 = \frac{r_2}{r_2 + R_5}*i$

$i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53$

$i_5 = 0.67 mA$

now the total current in R5 will be given by super position which is

$i = 0.67 + 0.18 = 0.85 mA$

so there is 0.85 mA current through R5 resistance

8 0

3 years ago

A piece of gum becomes stuck upon a skateboard's wheel. What is the centripetal acceleration of the piece of gum if the wheel's

Vilka [71]

Answer:

a

Explanation:

a_c = v_t^2/r

a_c = (0.5)^2/0.03

a_c = 8.33 m/s^2

7 0

3 years ago