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Verizon [17]
3 years ago
13

Which would be a common-sense practice in a lab environment?

Physics
1 answer:
algol133 years ago
6 0
Using safe research practices would be a very good common-sense practice in a lab environment.
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Harry and ron set up this experiment with a glider, whose mass they have measured to be 565 g, and seven washers hanging from th
svetlana [45]
Let's call m=565~g=0.565~kg the mass of the glider and m_w=7\cdot12~g =84~g=0.084~kg the total mass of the seven washers hanging from the string. 
The net force on the system is given by the weight of the hanging washers:
F_{net} = m_w g
For Newton's second law, this net force is equal to the product between the total mass of the system (which is m+m_w) and the acceleration a:
F_{net}=(m+m_w)a
So, if we equalize the two equations, we get
m_w g = (m+m_w)a
and from this we can find the acceleration:
a= \frac{m_w g}{(m+m_w)} = \frac{0.084~kg \cdot 9.81~m/s^2}{(0.565~kg+0.084~kg)}=1.27~m/s^2
5 0
3 years ago
A car driving at 20 m/s accelerates continuously 2m/s2​ ​ for 3 seconds. What is its final velocity
Lesechka [4]

Answer: V= u+ at

V= final velocity

u=initial velocity

a=acceleration

t=time taken

V= 20 + 2*3

V= 26m/s

Explanation:

5 0
2 years ago
What requirement must a force acting on a object satisfy in order for the object to undergo simple harmonic motion?
viktelen [127]

Answer:

Simple harmonic motion is the movement of a body or an object to and from an equilibrium position. In a simple harmonic motion, the maximum displacement (also called the amplitude) on one side of the equilibrium position is equal to the maximum displacement.

The force acting on an object must satisfy Hooke's law for the object to undergo simple harmonic motion. The law states that the force must be directed always towards the equilibrium position and also directly proportional to the distance from this position.

6 0
2 years ago
Your moods is approved when you have extra stress hormones in your blood
jenyasd209 [6]
Is this a true or false question?
5 0
3 years ago
A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac
Tpy6a [65]

Answer:

The force on one side of  the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

y=\dfrac{\Delta y}{\sin60}

y=\dfrac{2\Delta y}{\sqrt{3}}

The area of strip is

dA=\dfrac{9\times2\Delta y}{\sqrt{3}}

We need to calculate the force on  one side of  the plate

Using formula of pressure

P=\dfrac{dF}{dA}

dF=P\times dA

On integrating

\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}

F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}

F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})

F=3093529.3\ N

Hence, The force on one side of  the plate is 3093529.3 N.

7 0
3 years ago
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