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3 years ago

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A car travels at a constant speed up a ramp making an angle of 28 degrees with the horizontal component of velocity is 40 kmh^-1

, find the vertical and horizontal component of velocity.

1 answer:

Hoochie [10]3 years ago

7 0

Answer:

Vx = 35.31 [km/h]

Vy = 18.77 [km/h]

Explanation:

In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.

$v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h]$

In order to find the vertical component, we must use the sine function of the angle.

$V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h]$

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An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl

svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Displacement, s = 192 m

Acceleration, a = 9.81 m/s²

Substituting

s = ut + 0.5 at²

192 = 0 x t + 0.5 x 9.81 x t²

t = 6.26 seconds

Now we need to find horizontal distance traveled in 6.26 seconds by the package.

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 41 m/s

Time, t = 6.26 s

Acceleration, a = 0 m/s²

Substituting

s = ut + 0.5 at²

s = 41 x 6.26 + 0.5 x 0 x 6.26²

s = 256.52 m

The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

Initial velocity, u = 0 m/s

Time, t = 6.26 s

Acceleration, a = 9.81 m/s²

Substituting

v = u+ at

v = 0 + 9.81 x 6.26 = 61.41 m/s

Vertical component of velocity = 61.41 m/s

4 0

3 years ago

Two small metal spheres are 25 cm apaft.The spheres have equal amount of negative charge and repel each other with a force of 0.

Mars2501 [29]

Answer:

$0.5\times 10^{-6}C$

Explanation:

According to coulombs law force between two charges is given by $F=\frac{1}{4\pi \epsilon _0}\frac{Q_1Q_2}{R^2}$ here R is the distance between both the charges which is given as 25 cm

We have given force F =0.036 N

So $0.036=\frac{1}{4\pi \times 8.85\times 10^{-12}}\frac{Q^2}{(0.25^2)}$ As $\epsilon _0$ is constant which value is $8.85\times 10^{-12}$

$Q^2=0.250\times 10^{-12}$

$Q=0.5\times 10^{-6}C$

8 0

3 years ago

Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t

katrin2010 [14]

Answer:

$1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

23.4843749996 m

Yes

Explanation:

E = Electric field = $2\times 10^5\ N/C$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of proton= $1.67\times 10^{-27}\ kg$

q = Charge of electron = $1.6\times 10^{-19}\ C$

Acceleration is given by

$a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g$

The acceleration is $1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m$

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0

3 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Then the electric field at the point of interest is estimated as:

$E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}$

6 0

3 years ago

Indique onde em quantos metros o rapaz chegará com as seguintes condições: S0 = 0 a) v = 3 m/s e t = 2 s b) v = 2 m/s e t = 3,5

Naily [24]

Answer:

I will answer in English.

Here we will use the relation

Velocity*time = distance

So:

a) velocity = 3m/s

time = 2s

Distance = 3m/s*2s = 6m

b) velocity = 2m/s

time = 3.5s

Distance = 2m/s*3.5s = 7m

c) velocity = 10m/s

time = 0.5s

Distance = 10m/s*0.5s = 5m

d) velocity = 4m/s

time = 2.5s

Distance = 4m/s*2.5s = 9m

e) velocity = 1.5m/s

time = 5s

Distance = 1.5m/s*5s = 7.5m

8 0

3 years ago