Find the right answer please

What is the Difference between accuracy and precision ?<br>

Pie

Answer:

Accuracy measures how close results are to the true or known value. Precision, on the other hand, measures how close results are to one another.

6 0

1 year ago

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a 1500 kg car traveling at 15 m/s to the south collides with a 4500 kg truck that is intially at rest at a spotlight. The car an

harkovskaia [24]

Answer:

3.75 m/s south

Explanation:

Momentum before collision = momentum after collision

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Since the car and truck stick together, v₁ = v₂.

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v

Given m₁ = 1500 kg, u₁ = -15 m/s, m₂ = 4500 kg, and u₂ = 0 m/s:

(1500 kg) (-15 m/s) + (4500 kg) (0 m/s) = (1500 kg + 4500 kg) v

-22500 kg m/s = 6000 kg v

v = -3.75 m/s

The final velocity is 3.75 m/s to the south.

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2 years ago

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Two circular disks spaced 0.50 mm apart form a parallel-plate capacitor. transferring 1.50 ×109 electrons from one disk to the o

blondinia [14]

The diameter of the disks is 1.32 cm.

If n electrons each of charge e are transferred from one disc to another, calculate the total charge Q transferred from one disc to the other using the expression,

$Q=ne$

Substitute 1.50×10⁹ for n and 1.6×10⁻¹⁹C for e.

$Q=ne\\ =(1.50*10^9)(1.6*10^-^1^9C)\\ =2.4*10^-^1^0C$

The potential difference V between the disks separated by a distance d is given by,

$V=Ed$

here, E is the electric field.

Substitute 2.00×10⁵N/C for E and 0.50×10⁻³m for d.

$V=Ed\\ =(2.00*10^5N/C)(0.50*10^-^3m)\\ =100V$

The capacitance C of the capacitor is given by,

$C=\frac{Q}{V} \\ =\frac{(2.4*10^-^1^0C)}{100V} \\ =2.4*10^-^1^2F$

The capacitance of a parallel plate capacitor is given by,

$C=\frac{\epsilon_0A}{d}$

Here, ε₀ is the permittivity of free space and A is the area of the disks.

Rewrite the expression for A.

$C=\frac{\epsilon_0A}{d}\\ A=\frac{Cd}{\epsilon_0} \\ =\frac{(2.4*10^-^1^2F)(0.50*10^-^3m)}{(8.85*10^-^1^2C^2/Nm^2)} \\ =1.36*10^-^4m^3$

the area A of the disks is given by,

$A=\frac{\pi D^2}{4} \\ D=\sqrt{\frac{4A}{\pi } }$

Here, D is the diameter of the disk.

$D=\sqrt{\frac{4A}{\pi } }\\ =\sqrt{\frac{4(1.36*10^-^4m^2)}{3.14} } \\ =0.01316m\\ =1.32cm$

The diameter of each disc is found to be 1.32 cm.

7 0

2 years ago

A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon

Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force $F_p$ by the worker must be equal to the friction force $F_f$ on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

$F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N$

b. The work is done on the crate by this force is the product of its force $F_p$ and the distance traveled s = 4.35

$W_p = F_ps = 79.1*4.35 = 344 J$

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

$W_f = F_fs = -79.1*4.35 = -344 J$

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

$W_p + W_f = 344 - 344 = 0 J$