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3 years ago

6

A soccer ball kicked on a level field has an initial vertical velocity component of 15.0 mps. assuming the ball lands at the sam

e height from which it was kicked was the total time of the ball in the air?

1 answer:

telo118 [61]3 years ago

5 0

Answer:

The time taken for the ball to fly up in the air and back down again is 3.058 seconds.

Explanation:

Since the ball ends up at the same vertical distance ( on the ground) as it was at the start of its motion, we can set the total displacement of the ball equal to 0.

Thus, this problem can be simply solved by the following equation of motion:

$s = u*t + \frac{1}{2} (a*t^2)$

Here, s = total change in distance = 0 m

u = initial speed = 15 m/s

a = acceleration due to gravity = -9.81 m/s^2

t = time (to be found)

Substituting these values in the equation we get:

$0=15t+0.5(-9.81t^2)$

$-15t = -4.905t^2$

$t=15/4.905$

t = 3.058 seconds

So, the time taken for the ball to fly up in the air and back down again is 3.058 seconds.

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A weather station records the wind as blowing from the northeast at 12 km/h.

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vector quantities always have spedd and direction

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Mr. Beall paddles his canoe at 8.0 m/s North in a river that flows at 6.0 m/s to the South. What is the magnitude and direction

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Explanation:

Given that,

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We need to find the magnitude and the direction of the resulting velocity.

Let North is positive and South is negative. When two velocities are in opposite direction, they adds up. So,

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3 years ago

The distance between the earth and the sun is about 2.5x10¹¹m. Express SI prefix in kilometers

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Answer: 2.5x10^8 kilometers

Explanation:

Given that:

Distance between the earth and the sun = 2.5x10¹¹metres

Convert distance in metres to kilometers

Recall that 1000 metres = 1 kilometers

So, 2.5x10¹¹metres = Z kilometers

To get the value of Z, cross multiply

Z km x 1000 m = (2.5x10¹¹metres x 1km)

1000Z = 2.5 x 10^11

Divide both sides by Z

1000Z/1000 = 2.5 x 10^11/1000

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Thus, the distance between the earth and the sun is about 2.5x10^8km

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3 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

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4 years ago

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