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2 years ago

A 5. 3 ft -ft-tall girl stands on level ground. The sun is 30 ∘ above the horizon. How long is her shadow?

1 answer:

8 0

The trigonometric ratios are sine, cosine and tangent. We can use cosine to find the length of the shadow from the calculation as 1.7 ft.

<h3>What is trigonometric ratio?</h3>

The term trigonometric ratio has to do with sine, cosine and tangent which are used to solve problems that involve the right angled triangle.

Using the geometry of the right angle triangle, we can find the tangent of the angle 30 in order to obtain the length of the shadow.

Tan 30° = opp/3

opp = 3Tan 30°

opp = 1.7 ft

The length of the girl's shadow from the calculation is 1.7 ft.

Learn more about tangent: brainly.com/question/14022348

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A rigid body rotates about a fixed axis with a constant angular acceleration. Which one of the following statements is true conc

Korolek [52]

To solve this problem it will be necessary to apply the concepts related to angular acceleration and tangential acceleration. Definitions are given in the description of the angular kinematic movement and describe these two expressions of acceleration as:

The angular acceleration of the object is written as,

$\alpha = \frac{\Delta \omega}{\Delta t}$

The tangential acceleration is

$a_T = \alpha r$

Therefore equating the two previously expression we have,

$a_T = \frac{\Delta \omega}{\Delta t} r$

Here we can conclude that the angular acceleration depends on the change in angular velocity.

Therefore the correct answer is B.

5 0

3 years ago

A net force of 5N is applied to a 13kg mass. what is its acceleration?

liq [111]

Answer:

$\boxed {\boxed {\sf a \approx 0.4 \ m/s^2}}$

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

$F= m \times a$

The net force is 5 Newtons and the mass is 13 kilograms. Let's convert the units for force to make the problem and unit cancellation easier.

1 Newton (N) is equal to 1 kilogram meter per square second (1 kg*m/s²)
The net force of 5 N is equal to 5 kg*m/s²

Now we know the values for 2 variables:

F= 5 kg*m/s²
m= 13 kg

Substitute the values into the formula.

$5 \ kg *m/s^2 = 13 \ kg * a$

Since we are solving for the accleration we must isolate the variable, a. It is being multiplied by 13 kilograms and the inverse of multiplication is division. Divide both sides by 13 kg

$\frac {5 \ kg *m/s^2}{13 \ kg}= \frac{ 13 \ kg *a}{13 \ kg}$

$\frac {5 \ kg *m/s^2}{13 \ kg}=a$

The units of kilograms (kg) cancel.

$\frac {5 m/s^2}{13 }=a$

$0.384615385\ m/s^2=a$

The original measurements of force and mass ( 5 and 13) have 1 and 2 significant figures. We must round our answer to the least number of sig figs: 1.

For the number we found, that is the tenths place. The 8 in the hundredth place (0.384615385) tells us to round the 3 up to a 4.

$0.4 \ m/s^2 \approx a$

The acceleration is approximately <u>0.4 meters per square second.</u>

5 0

3 years ago

According to car and driver, an alfa romeo going at 70 mph requires 177 feet to stop. assuming that the stopping distance is pro

tia_tia [17]

Let d represents the stopping distance required by an alfa romeo.

The stopping distance is proportional to the square of velocity is equivalent to the equation,

$d=kv^2$

Here, k is proportionality constant.

Given, d = 177 feet and v = 70 mph.

So, proportionality constant

$k=\frac{177\ feet }{(70\ mph)^2} \simeq 0.0361$ .

Thus, the stopping distances required by an alfa romeo going at 60 mph and at 140 mph,

$d_{60mph} =0.0361 \times (60\ mph)^2 =129.96\ feet$

and

$d_{140mph} =0.0361 \times (140\ mph)^2 =707.56\ feet$

3 0

3 years ago

A charge of 6.65 mC is placed at each corner of a square 0.500 m on a side.

mixer [17]

Answer:

Explanation:

Given that:

length of side , a = 0.5 m

charge , q = 6.65 mC

length of diagonal , d = 0.5 * sqrt(2)

d = 0.707 m

F is the force due to adjacent particle ,

F1 is the force due to diagonal particle

Now , for the net charge on a particle

Fnet = 2 * F * cos(45) + F1

Fnet = 2*cos(45) * k * q^2/a^2 + k * q^2/d^2

Fnet = 9*10^9 * 0.00665^2 * (2* cos(45)/.5^2 + 1/.707^2)

Fnet = 3.05 *10^6 N

the magnitude of net force acting on each particle is 3.05 *10^6 N

part B)

for the direction of particle

d) along the line between the charge and the center of the square outward of the center

7 0

3 years ago

Bicycle A with mass 40 Kg is traveling with a velocity of 4 m/s and Bicycle B with mass 20 Kg is traveling with a velocity of 2m

andrew-mc [135]

Answer:

bicycle A has a greater K.E.

Explanation:

K.E = 1/2mv²

bicycle A = 1/2 × 40 × 4² = 320J

bicycle B = 1/2 × 20 × 2² = 100J

bicycle A has a greater K.E. because it has bigger mass and moves with faster velocity

7 0

3 years ago