If a truck weighs 195,000N what is the mass of the truck

A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with constant speed of 3.50m/s. The coefficient of kin

stellarik [79]

Answer:

3.125 m

Explanation:

We are given that

Mass of box=m=11.2 kg

Speed of box=u=3.5m/s

Coefficient of kinetic friction= $\mu_k=0.2$

Final velocity,v=0

a.We have to find the horizontal force applied by worker to maintain the motion.

According to question

Horizontal force=F= $f=\mu_kmg$

$g=9.8m/s^2$

Substitute the values

Horizontal force= $F=0.2\times 11.2\times 9.8=21.95 N$

b.According to work-energy theorem

$W=\frac{1}{2}mv^2-\frac{1}{2}mu^2$

$-\mu mg s=\frac{1}{2}(11.2)(0)^2-\frac{1}[2}(11.20)(3.5)^2$

$-\mu mg s=-\frac{1}{2}(11.2)(3.5)^2$

$0.2\times (11.2)\times 9.8\times s=\frac{1}{2}(11.2)(3.5)^2$

$s=\frac{11.2\times (3.5)^2}{2\times 0.2\times 11.2\times 9.8}$

$s=3.125 m$

Hence, the box slide before coming to rest=3.125 m

4 0

3 years ago

The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 651 nm in vacuum. This light propag

sukhopar [10]

Answer:

$v=2.58\times10^8m/s$

Explanation:

The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or $n=\frac{c}{v}$ . For our case we want to use $v=\frac{c}{n}$ , which for our values is equal to:

$v=\frac{c}{n}=\frac{299792458m/s}{1.16}=258441774.138m/s$

Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the <em>least</em> number of significant figures):

$v=2.58\times10^8m/s$

4 0

3 years ago

Two long, straight parallel wires are placed 38 cm apart, one above the other. The top and bottom wires are carrying currents 4.

ElenaW [278]

Answer:

The force per unit length (N/m) on the top wire is 16.842 N/m

Explanation:

Given;

distance between the two parallel wire, d = 38 cm = 0.38 m

current in the first wire, I₁ = 4.0 kA

current in the second wire, I₂ = 8.0 kA

Force per unit length, between two parallel wires is given as;

$\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d }$

where;

μ₀ is constant = 4π x 10⁻⁷ T.m/A

Substitute the given values in the above equation and calculate the force per unit length

$\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d } = \frac{4\pi *10^{-7}*4000*8000 }{2\pi *0.38} = 16.842 \ N/m$

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m

4 0

3 years ago

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3, length 88.8 cm and diameter 2.30 cm fro

vichka [17]

Answer:

w = 28.25 N

Explanation:

To do this, we need to use two expressions.

First, to calculate the weight of any object, we use the 2nd law of newton. In this case, the weight is:

w = m*g (1)

However we do not have the mass of the rod. We need to calculate that. To calculate the mass, we'll use the expression of density which is:

d = m/V

From here, we solve for mass:

m = d * V (2)

Finally, we can know the volume of the rod, because is cylindrical, therefore, the volume of a cylinder is:

V = π * r² * h (3)

So, in resume, we need to solve for the volume of the rod, then, the mass ans finally the weight. Let's calculate the volume of the rod, converting the units of centimeter to meters, just dividing by 100:

diameter = 2.3 cm ---> radius = 2.3/2 = 1.15 cm -----> 0.0115 m

Length or height = 88.8 cm ----> 0.888 m

Replacing in (3):

V = π * (0.0115)² * 0.888

V = 3.69x10⁻⁴ m

Now, let's use (2) to calculate the mass:

m = 7800 * 3.69x10⁻⁴

m = 2.88 kg

Finally for the weight, we'll use expression (1):

w = 2.88 * 9.81

<h2>w = 28.25 N</h2><h2>And this is the weight of the rod.</h2>

4 0

3 years ago

A thin film of cryolite ( nc = 1.34 ) is applied to a camera lens ( ng = 1.58 ). The coating is designed to reflect wavelengths

rosijanka [135]

Answer:

=99.07nm

Explanation:

minimum thickness

2nd = (m - 1/2)λ

d = (m - 1/2)(λ/2n)

refractive index of the thin film, n = 1.34

minimum thickness m = 1

light wavelength λ = 531nm

d = (1 - 1/2) (531 / (2)(1.34)

d = 531/5.36

= 99.07nm

3 0

3 years ago