What is the acceleration of the object? m/s2

What is an ampere? NO GUESSING! ANSWERS ONLY!

Igoryamba

Answer:

the SI base unit of electrical current

7 0

3 years ago

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Why did boy (the observer) hear the pitch of the sound change? What does this tell us about the frequency of the sound? Support

timofeeve [1]

Answer:

because of the doppler effect

Explanation:

Remember, the doppler effect refers to the changes in sound (frequency of sound) observed by a person who is in a position relative to the wave source.

In this example, we notice as the train comes closer to the boy, the sound becomes louder also increasing the pitch slightly, the doppler effect sets in when the train passes the boy because the boy notices a decrease in the pitch of the moving train.

We learn from the change in the observed sound of the train that the frequency of the sound is determined by the distance of the observer from the wave source.

In other words, the closer the source of the sound to the observer; the faster it travels to the observer, however, the farther it is; the lesser it is; the greater the sound heard.

5 0

3 years ago

A baseball pitcher throws a fastball horizontally at a speed of 43.0 m/s. Ignoring air resistance, how far does the ball drop be

Nady [450]

Based on your problem where as ask for the distance of the ball drop between the pitchers mound and the home plate and with a given of the speed of ball is 43m/s and the homeplates is 60.6ft away. Based on my step by step procedure and also considering the value of gravity by 9.8m/s^2 i came up with the distance of 144m away

7 0

3 years ago

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A 55 kg person falling with a velocity of 0.6

siniylev [52]

Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

K.E= kinetic energy

M=mass

V=velocity

K.E =0.5*55*0.6^2

K.E=9.9J

Explanation:

3 0

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 $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

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

6 0

3 years ago