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3 years ago

How is space exploration using space shuttles similar to and different from space exploration using space stations

Physics

1 answer:

solniwko [45]3 years ago

4 0

Its simlar because they both are studying space and its diffrence because one is u actually living it

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Darius' boat sails into the harbor with a speed of 80m/s. After 20 seconds, Darius' boat has come to a stop at the dock. What is

Novosadov [1.4K]

Answer:

The boat's acceleration is 4 m/s²

Explanation:

This question seeks to test the knowledge of acceleration and how to calculate acceleration when the speed is provided. Hence, the formula for acceleration would be used here.

Acceleration (m/s²) = speed (in m/s) ÷ time (in seconds)

Acceleration = 80 m/s ÷ 20 secs

Acceleration = 4 m/s² or 4 ms⁻²

The boat's acceleration is 4 m/s²

4 0

3 years ago

A 2 N and an 8 N force pull on an object to the right and a 4 N force pulls on the same object to the left. If the object has a

Elden [556K]

Answer:

a = 12 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

ΣF = m*a

where:

ΣF = sum of forces acting on a body [N] (units of Newtons)

m = mass = 0.5 [kg]

a = acceleration [m/s²]

Let's take the direction of positive forces to the right and negative forces directed to the left

2 + 8 - 4 = 0.5*a

6 = 0.5*a

a = 12 [m/s²]

7 0

3 years ago

Question 10 (1 point)

torisob [31]

Answer:

It does not hit the students face because the speed of the balloon slows down as energy is lost through thermal.

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

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