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3241004551 [841]
3 years ago
11

TRUE or FALSE: An object's mass changes depending on which planet it is located.

Physics
1 answer:
Vedmedyk [2.9K]3 years ago
7 0

Answer:

Fales

Explanation:

That's why scientists and engineers often measure an object's mass—how much matter the object contains—rather than its weight. Mass stays the same regardless of location and gravity. You would have the same mass on Mars or Jupiter as you do here on Earth. Your weight is different on other planets due to gravity.

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Find B when θ=35, E=10V, t=5, N=250, A=1.20m^2
givi [52]

Answer:

its most definitely c. trust me

Explanation:

3 0
3 years ago
A rock is thrown horizontally from a building at 15 m/s. It hits the ground 45 m from the base of the building. How high was the
evablogger [386]

Answer:

Explanation:

hope this helps!

7 0
3 years ago
An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce
Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Frequency of pendulum is given by

f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}

Given in the question

f_p=\dfrac{1}{2}f_s

\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m

The unstretched length of the spring is 2.67 m

6 0
3 years ago
If a 5.5 kg object experiences 15 N of force for .15 seconds what is the speed change
Nady [450]

The speed change : Δv = 0.41 m/s

<h3>Further explanation</h3>

Given

mass = 5.5 kg

Force = 15 N

time = 0.15 s

Required

the speed change

Solution

Newton 2nd's law

Impulse and momentum

F = m.a

F = m . Δv/t

F.t = m.Δv

Input the value :

15 N x 0.15 s = 5.5 kg x Δv

Δv = 0.41 m/s

8 0
3 years ago
Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),
salantis [7]

Answer:

The  drag coefficient is  D_z  =  1.30512  

Explanation:

From the question we are told that

     The density of air is  \rho_a  = 1.21 \ kg/m^3

     The diameter of bottom part is  d = 0.15 \ m

The  power trend-line  equation is mathematically represented as

      F_{\alpha }  = 0.9226 * v^{0.5737}

let assume that the velocity is  20 m/s

Then

      F_{\alpha }  = 0.9226 * 20^{0.5737}

       F_{\alpha }  = 5.1453 \ N

The drag coefficient is mathematically represented as

      D_z  =  \frac{2 F_{\alpha } }{A \rho v^2 }

Where  

     F_{\alpha } is the drag force

      \rho is the density of the fluid

       v is the flow velocity

       A is the area which mathematically evaluated as

       A = \pi r^2 =  \pi  \frac{d^2}{4}

substituting values

     A =  3.142 *    \frac{(0.15)^2}{4}

     A = 0.0176 \  m^2

Then

   D_z  =  \frac{2 * 5.1453 }{0.0176 * 1.12 *  20^2 }

   D_z  =  1.30512  

3 0
3 years ago
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