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2 years ago

Help me with this please

Physics

1 answer:

lara31 [8.8K]2 years ago

7 0

Answer:

for the secound one i think it would be 45.87 let me know if this is wrong I love to get my feed back

Explanation:

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the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. the s

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When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat.

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3 years ago

A waitperson carrying a tray with a platter on it tips the tray at an angle of 12° below the horizontal. If the gravitational fo

Nadusha1986 [10]

Answer:

The answer is 1.0 N

Explanation:

inclination of tray=12^{\circ}

gravitational Force=5 N

Now this gravitational force has two component i.e.

5\sin \theta is parallel to the tray =1.039 N

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2 years ago

A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing a

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Answer:

5 m/s2

Explanation:

The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.

The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle

$a_c = \frac{v^2}{r} = \frac{30^2}{300} = \frac{900}{300} = 3 m/s^2$

So the magnitude of the total acceleration is

$a = \sqrt{a_c^2 + a_l^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 m/s^2$

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3 years ago

An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be

marysya [2.9K]

Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75 in +ve x- direction

$F_1 = \dfrac{GM_1 m}{R_1^2}$

$F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}$

$F_1 = 2.019\times 10^{8}\ m$

Force applied due to mass 14 Kg in -ve x-direction

$F_2 = \dfrac{GM_2 m}{R_2^2}$

$F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}$

$F_2 = 0.857\times 10^{8}\ m$

net force

F = F₁ + F₂

$F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m$

$F = 1.162\times 10^{8}\ m$

Using newton second law

$a = \dfrac{F}{m}$

$a = \dfrac{ 1.162\times 10^{8}\ m}{m}$

$a =1.162\times 10^{8} \ m/s^2$

b) As the acceleration of mass comes out to be +ve hence, the direction will be toward the mass of 8.75 Kg

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3 years ago

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Answer:

There is pressure on both the in and outsides of the balloon: high inside and low outside, respectively. In terms of air mass, high pressure has greater activity and tends to move upward (balloon rises); whereas low presssure, being slower, descends (balloon descends).

Explanation:

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2 years ago

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