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lara31 [8.8K]
3 years ago
15

"determine the resultant internal loadings acting at the cross sections at points f and g of the frame. set θ = 27º and t = 178

lb."

Physics
1 answer:
Leviafan [203]3 years ago
5 0

Hi you didn't provide any images to solve the question, hence I am going to solve a different question of same concept so you can have an idea how to tackle such types of questions.(please refer to the attachment for question)

Answer:

<u> Please refer to the attachment for answers and explanation</u>

Explanation:

<u> Please refer to the attachment for answers and explanation</u>

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Choose the best real world exzample of convection:
spin [16.1K]
The correct answer is A
3 0
2 years ago
Read 2 more answers
many lichens are composed of fungi and algae. The fungi get sugars from the algae, and the algae get water, minerals, and protei
Tcecarenko [31]

Answer:

The answer is mutualism because they are both on the receiving and giving ends

6 0
3 years ago
A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f
zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s

v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

s = v_vt + gt^2/2

10 = 19.97t - 9.8t^2/2

4.9t^2 - 19.97t + 10 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}

t= \frac{19.9658877511823\pm14.24}{9.8}

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

s_1 = v_ht_1 = 15*0.58 = 8.8 m

s_2 = v_ht_2 = 15*3.49 = 52.5m

8 0
2 years ago
Which of the following statements is TRUE about impulse? (Consider the Impulse-Momentum Theorem: F▲t = m ▲v )
kotykmax [81]

The statement “Impulse is a vector quantity” is true about Impulse.

Answer: Option B

<u>Explanation: </u>

The object’s action by applied force in a particular time interval, there happens changing in momentum called impulse. It is denoted by a symbol ‘J’ or ‘imp’ and expressed in a unit ‘Ns’. As impulse depends on the acted force, when a collision arises from front, behind or side, the force’s direction would be differed.

                   \text {Impulse }=\text {Force } \times \text {time}=\vec{F} \Delta t

So, from this option A is false as impulse is not a force but changing momentum. The unit is not Newton, it is Newton second (Ns). The force direction differs (impulse direction) for each cases of collision, so option D also false. Hence, option B seems to be correct. Vector quantity deals with both direction and magnitude and important in motion study.

8 0
3 years ago
How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances
Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = x+y=149.6\times 10^6\ m

G = Gravitational constant

M_s = Mass of Sun = 1.989\times 10^{30}

M_e = Mass of Earth = 5.972\times 10^{24}\ kg

According to the question

\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y

x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m

The probe should be 258774.9441 m from Earth

7 0
2 years ago
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