Answer:
(1,)
Explanation:
Answer:
magnitude: 21.6; direction: 33.7°
Explanation:
A chef places an open sack of flour on a kitchen scale. The scale reading of
40 N indicates that the scale is exerting an upward force of 40 N on the sack. The magnitude of this force equals the magnitude of the force of Earth’s gravitational attraction on the sack. The chef then exerts an upward force of
10 N on the bag and the scale reading falls to 30 N.Draw a free-body diagram of the latter situation.
Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
Answer:
40N
Explanation:
trust
waht is science
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Answer:
the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.
Explanation:
Two protons are a distance 3 10-9 m apart. What is the electric potential energy of the system consisting of the two protons
Answer:
The electric potential energy of the system is 7.87x10⁻²⁰ J.
Explanation:
The electric potential energy is given by:
[tex]E = \int{Fdr} = \frac{Kq_{1}q_{2}}{r}[/tex]
Where:
q₁ = q₂ is the charge of the protons = 1.62x10⁻¹⁹ C
r is the distance = 3x10⁻⁹ m
K: is the electrostatic constant = 9x10⁹ Nm²/C²
[tex] E = \frac{Kq_{1}q_{2}}{r} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*(1.62 \cdot 10^{-19} C)^{2}}{3\cdot 10^{-9} m} = 7.87 \cdot 10^{-20} J [/tex]
Therefore, the electric potential energy of the system is 7.87x10⁻²⁰ J.
I hope it helps you!
The electric potential energy of the system should be 7.87x10⁻²⁰ J.
Calculation of the electric potential energy:SInce We know that
fdr = kq1q2/r
Here
q₁ = q₂ i.e. is the charge of the protons = 1.62x10⁻¹⁹ C
r should be the distance = 3x10⁻⁹ m
K should be the electrostatic constant = 9x10⁹ Nm²/C²
Now electric potential energy should be
= (9x10⁹ Nm²/C² * 1.62x10⁻¹⁹ C) / 3x10⁻⁹ m
= 7.87x10⁻²⁰ J.
hence, The electric potential energy of the system should be 7.87x10⁻²⁰ J.
learn more about energy here: https://brainly.com/question/17384612
Please provide explanation!!!
Thank you.
Answer:
(a) 102 cm/s
(b) 0.490 cm²
Explanation:
(a) Use Bernoulli equation.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0
½ ρ v₁² + ρgh₁ = ½ ρ v₂²
½ v₁² + gh₁ = ½ v₂²
½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²
v = 102 cm/s
(b) The flow rate is constant.
v₁ A₁ = v₂ A₂
(25.0 cm/s) (2.00 cm²) = (102 cm/s) A
A = 0.490 cm²
How do I proton and and electron compared
block of mass m sits at rest on a rough inclined ramp that makes an angle with the horizontal. What must be true about normal force F on the block due to the ramp
Answer:
Explanation:
For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.
The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction
Taking the sum of forces along the y component
Sum Fy = -W+R = ma
Since acceleration is zero
-W+R = m(0)
-W+R = 0
-W = -R
W = R
Hence the Normal reaction force acting on the on the body is equal to normal force