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3 years ago

6

Which of these activities increases the amount of carbon in the atmosphere? A. animals eating plants B. burning of fossil fuels

C. fossilization D. photosynthesis

1 answer:

tester [92]3 years ago

8 0

Answer:

B. Burning of fossil fuels

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A cannon fires a 0.652 kg shell with initial

laila [671]

Mass have no effect for the projectile motion and u want to know the height "h"
first,
find the vertical and horizontal components of velocity
vertical component of velocity = 12 sin 61
horizontal component of velocity = 12 cos 61
now for the vertical motion ;
S = ut + (1/2) at^2
where
s = h
u = initial vertical component of velocity
t = 0.473 s
a = gravitational deceleration (-g) = -9.8 m/s^2

h=[12×sin 610×0.473]+[−9.8×(0.473)2]

u can simplify this and u will get the answer

h=.5Gt2

H=1.09m

6 0

3 years ago

Read 2 more answers

To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv

Lesechka [4]

Answer:

E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

λ = Q / L

If we derive from the length we have

λ = dq/dx ⇒ dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

dE = k dq / x²2

dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

E = k $\int\limits^{d+L}_d {\lambda/x^{2}} \, dx$

We take out the constant magnitudes and perform the integral

E = k λ (-1/x) ${(-1/x)}^{d+L} _{d}$

Evaluating

E = k λ [ 1/d - 1/ (d+L)]

Using λ = Q/L

E = k Q/L [ 1/d - 1/ (d+L)]

let's use a bit of arithmetic to simplify the expression

[ 1/d - 1/ (d+L)] = L /[d(d+L)]

The final result is

E = k Q / [d(d+L)]

3 0

3 years ago

As a projectile falls, what happens to the components of velocity?

netineya [11]

Answer:

Option (c).

Explanation:

An object when when projected at an angle, will have some horizontal velocity and vertical velocity such that,

$v_x=v\cos\theta\ \text{and}\ v_y=v\sin\theta$

$\theta$ is the angle of projection

The horizontal component of the projectile remains the same because there is no horizontal motion. Vertical component changes at every point.

As a projectile falls, vertical velocity increases in magnitude, horizontal velocity stays the same .

7 0

3 years ago

When the force acting on the body equal to acceleration?

topjm [15]

Answer:

Acceleration and velocity Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force.

5 0

2 years ago

A spring has a force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is

kenny6666 [7]

Answer:

$f2/f1 = \sqrt{2}$

Explanation:

From frequency of oscillation

$f = 1/2pi *\sqrt{k/m}$

Initially with the suspended string, the above equation is correct for the relation, hence

$f1 = 1/2pi *\sqrt{k/m}$

where k is force constant and m is the mass

When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

$f2 = 1/2pi *\sqrt{2k/m}$

Employing f2/ f1, we have

$f2/f1 = \sqrt{2}$

3 0

3 years ago