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sveticcg [70]
2 years ago
7

A 165 N object is supported by three cables(T1, T2 and T3), of which T1 and T2 are making angles θ1 = 52o and θ2 = 39o as shown

in Figure.
Here T3 = 165 N, calculate the tensions T1 and T2.

1


(a) Tension in string 1, T1 (in Newton) =
Answer for part 1


(b) Tension in string 2, T2 (in Newton) =
Answer for part 2
Hhgeppp plzzz​

Physics
1 answer:
Liono4ka [1.6K]2 years ago
3 0

Answer:

?Tension in string 2, T2 (in Newton) =

Answer for part 2

Hhgeppp plzzz

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A 57 kg person in a rollercoaster moving through the bottom of a curved track of radius 42.7 m feels a normal force of 995 N. Ho
natita [175]

Answer:

Use Fc centripetal force as positive and W the weight as negative

N = m v^2 / R + m g

v^2 = (N - m g) R / m

v^2 = (995 - 57 * 9.8) 42.7 / 57 = 327 m^2/s^2

v = 18.1 m/s

Note: N - m g is the net force producing the centripetal force

5 0
2 years ago
What is the primary force that causes the seafloor to spread and continents to drift?
jekas [21]
The primary force would be Thermal Convection, it pushes it and thus makes continents drift.
-Hope that helps, Nexxmexx :3
5 0
3 years ago
What happen to the frequency of a sound wave as you change the pitch?
steposvetlana [31]

The "pitch" of a sound is the impression your brain forms
that corresponds to the frequency of the sound wave.

When the frequency is high, your brain says "high pitch".

When the frequency is low, your brain says "low pitch".

4 0
3 years ago
You push on a box with 100 N of force, causing it to accelerate at 5 m/s?.
Xelga [282]
<h3><u>Given</u><u>:</u><u>-</u></h3>

Force,F = 100 N

Acceleration,a = 5 m/s²

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the mass of the box .

<h3><u>Formula</u><u> </u><u>used:-</u><u> </u></h3>

\bf \: Force = Mass  \times  Acceleration

<h3><u>Solution:-</u><u> </u></h3>

\sf \: Force = Mass × Acceleration

★ Substituting the values in the above formula,we get:

\sf \implies \: 100 = Mass \times 5

\sf \implies \: Mass =   \cancel\dfrac{100}{5}

\sf \implies \: Mass = 20 \: kg

4 0
3 years ago
Read 2 more answers
To stop a car, first you require a certain reaction time to begin braking; then the car slows under the constant braking deceler
ANEK [815]

Answer:

a) t_r = 0.55 s

b) a = 5.59 m/s²

Explanation:

given,

total distance traveled by the car to stop is 56.9 m when speed of vehicle is 80 km/h or 80 × 0.278 = 22.24 m/s

total distance traveled by the car to stop is 25.7 m when speed of vehicle is 50.7 km/h or 50.7 × 0.278 = 14.09 m/s

using stopping distance formula

s_1 = v_1 t_r +\dfrac{v_1^2}{2 a}................(1)

s_2 = v_2 t_r +\dfrac{v_2^2}{2 a}..............(2)

on solving both the equation we get

a = \dfarc{v_1v_2(v_1-v_2)}{2(s_1v_2-s_2v_1)}

a = \dfarc{22.4\times 14.09(22.24-14.09)}{2(56.9\times 14.09-25.7\times 22.24)}

a = 5.59 m/s²

now reaction time calculation

t_r =\dfrac{v_1^2d_2-v_2^2d_1}{v_1v_2(v_1-v_2)}

t_r =\dfrac{22.24^2\times 25.7-14.09^2\times 56.9}{22.4\times 14.09(22.24-14.09)}

t_r = 0.55 s      

8 0
3 years ago
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