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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 street lamp weighs 150N. It is supported by two wires that form an angle of 120° with each other. The tensions in each wire ar

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Answer:

so you take 120÷2 wires

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2 years ago

A 10.0 kg weather rocket generates a thrust of 230 NN . The rocket, pointing upward, is clamped to the top of a vertical spring.

blondinia [14]

Answer: 0.2m

Explanation: Firstly only the Rocket's Weight Compress the spring which can be found by

$F_r=M_r*g\\F_r=10*9.81\\F_r=98.1N$

According to Hooks Law

$F_r=k*x\\x=F_r/k\\x=98.1/480\\x=0.2m$

The part b and c of this question is done in the attachment

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3 years ago

A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on

kari74 [83]

Answer:

$9.16rad/s^2$

Explanation:

We are given that

Mass, $m_1=3.3 kg$

Radius,r=0.8 m

$m_2=4.9 kg$

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

$4.9g-T=4.9a$

$Tr=I\alpha$

Where

$\alpha=\frac{a}{r}$

$Tr=\frac{1}{2}m_1ra$

$T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a$

Substitute the value

$4.9g-\frac{1}{2}(3.3a)=4.9a$

$4.9\times 9.8=4.9a+\frac{3.3a}{2}$

Where $g=9.8 m/s^2$

$48.02=a(4.9+1.65)=6.55a$

$a=\frac{48.02}{6.55}=7.33m/s^2$

Angular acceleration, $\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2$

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3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

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2 years ago

Heather drops a ball weighing 0.50kg a distance of 2.0m. If the speed of the ball right after it bounces is the same speed of th

ki77a [65]

The speed of the ball just before impact was v=√(2gh) = 6.26m/s. The acceleration is twice this over the time (twice because the second speed is the same in the other direction, meaning the total change in speed is 2V)
a = 12.52/0.10 = 125.2m/s²
The force is F=ma, so F = 0.5kg·125.2m/s² = 62.6N

7 0

3 years ago