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3 years ago

1.) A ball is thrown horizontally from the top of a building. What is the vertical velocity of the ball after 1 second?

Physics

2 answers:

MatroZZZ [7]3 years ago

7 0

Answer:

B. 9.8 m/s
A. 0 m/s²
C. 45°
B. 70°
D. uniformly accelerated motion

Explanation:

u=0m/s

v=?

a=9.8m/s²

t=1sec

now

we have

v=u+at

v=0+9.8*1

v=9.8m/s

angle is measured in sin and cos angle where maximum angle's value is gained by 45°

we have

range

R= $\frac{u²sin2\theta}{g}$

For

angle =20°

R1= $\frac{u²sin2*20}{g}$ .......[1]

another angle be $\theta$

R2= $\frac{u²sin2\theta}{g}$ .......[2]

since R1=R2

By comparing we get

Sin40=Sin( $2\theta$ )

Sin(180-40)=Sin( $2\theta$ )

140= $2\theta$

$\theta$ =70°

BARSIC [14]3 years ago

6 0

9.8
zero
45
not sure - 70
Uniform accelerated motion

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2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

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$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

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Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

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So:-

$\dashrightarrow \sf work = force \times distance$

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$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

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$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

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$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

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$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

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$\dashrightarrow \bf work =\red{2250\: joule}$

5 0

2 years ago

Recall that photosynthetic rates remain relatively constant in regions near the equator. Imagine that tropical environments pers

solniwko [45]

Answer:E.

a straight line sloping upward (atmospheric CO2 levels would not seasonally oscillate, but would have increased over time)

Explanation: Photosynthesis is the process through which green plants and certain other organisms make their food using carbon dioxide and water in the he presence of Sunlight.

During photosynthesis, energy conversion from light energy into chemical energy takes place.

Photosynthesis is one of the major processes that helps to eliminate and reduce the quantity of carbon dioxide in the atmosphere,it also serve to make oxygen present in the atmosphere.

8 0

3 years ago

figure 2 shows a charged ball of mass m = 1.0 g and charage q = -24*10^-8 c suspended by massless string in the presence of a un

Vlad [161]

Answer:

E = 307667 N/C

Explanation:

Since the object's mass is 1 g, then its weight in newtons is 0.001 * 9.8 = 0.0098 N.

This weight should have the same magnitude of the vertical component of the tension T of the string (T * cos(37)) so we can find the magnitude of the tension T via:

0.0098 N = T * cos(37)

then T = 0.0098/cos(37) N = 0.01227 N

Knowing the tension's magnitude, we can find its horizontal component:

T * sin(37) = 0.007384 N

and now we can obtain the value of the electric field since we know the charge of the ball to be: -2.4 * 10^(-8) C:

0.007384 N = E * 2.4 * 10^(-8) C

Then E = 0.007384/2.4 * 10^(-8) N/C

E = 307667 N/C

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3 years ago

We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,

larisa [96]

Answer:

The final charges of each sphere are: q_A = 3/8 Q , q_B = 3/8 Q , q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

q_A = Q / 2

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Now sphere A touches sphere C, ending with half the charge

q_A = ½ (Q / 2) = ¼ Q

q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

q_A = 3/8 Q

q_B = 3/8 Q

The final charges of each sphere are:

q_A = 3/8 Q

q_B = 3/8 Q

q_C = 3/4 Q

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3 years ago

What role did gravity play in the formation of the planets?

joja [24]

Your answer would be D.
If an object has mass, it has gravity, and the more mass it has, the stronger its gravity. During the formation of planets, essentially, various matter and elements pulled and fused together (because of the gravity), forming planetesimals.

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3 years ago

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