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2 years ago

A force of 9.00 newtons acts at an angle of 19.0° to the horizontal. What is its component in the horizontal direction?

2 answers:

5 0

It would be helpful if you draw the figure of the problem. You will see that a right triangle would be constructed by the problem where 19.0 is the angle between the hypotenuse and the base of the triangle. It is said that the force acting is said to be 9.0 N at the said angle to the horizontal. Using trigonometric relations,

cos 19 = adjacent / hypotenuse = horizontal component / 9
horizontal component = 8.51 N

hoa [83]2 years ago

5 0

Answer:

8.51 N

Explanation:

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Consider the Bohr energy expression (Equation 30.13) as it applies to singly ionized helium He+ (Z = 2) and an ionized atom with

ella [17]

Answer:

Explanation:

Bohr's energy expression is as follows

E_n = 13.6 z² /n² where z is atomic no and n is principal quantum no of the atom .

z for helium is 2 and for ionised atom is 5 . Let energy of n₁ level of He is equal to energy level n₂ of ionised atom

13.6 x 2² / n₁² = 13.6 x 5² / n₂²

n₁ / n₂ = 2/5 , ie 2nd energy level of He matches 5 th energy level of ionised atom .

For quantum numbers less than or equal to 9 , If we take n₁ = 8 for He

Putting it in the equation above

2² / 8² = 5² / n₂²

n₂ = 5 x 8 / 2

= 20 .

energy

= - 13.6 x2² / 8²

= - 0.85 eV .

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2 years ago

Think about how geothermal energy is captured and used. Explain how geothermal energy shows the flow of thermal energy from hot

kumpel [21]

Answer:

People can capture geothermal energy through: Geothermal power plants, which use heat from deep inside the Earth to generate steam to make electricity. Geothermal heat pumps, which tap into heat close to the Earth's surface to heat water or provide heat for buildings

When the weather is cold, the water or refrigerant heats up as it travels through the part of the loop that's buried underground. Once it gets back above ground, the warmed water or refrigerant transfers heat into the building. The water or refrigerant cools down after its heat is transferred.

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2 years ago

A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp

liq [111]

Answer:

$W_2=\sqrt{\frac{3}{5} }W_1$

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

$I_1 =\frac{2}{5}MR^2$

$K_1 = \frac{1}{2}IW_1^2$

So, replacing, we get that:

$K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

At the same way, the moment of inertia and kinetic energy for second ball is:

$I_2 =\frac{2}{3}MR^2$

$K_2 = \frac{1}{2}IW_2^2$

So:

$K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2$

Then, $K_2$ is equal to $K_1$ , so:

$K_2 = K_1$

$\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2$

$\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2$

$\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2$

Finally, solving for $W_2$ , we get:

$W_2=\sqrt{\frac{3}{5} }W_1$

5 0

2 years ago

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0

2 years ago

Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr

Aloiza [94]

Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

$F=ma$

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

8 0

2 years ago

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