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Novosadov [1.4K]

3 years ago

7

and area' Explain the following in your answers, use the words pressure', 'force A truck used in the desert has wide tyres. 6 A

cricket stump has a sharp point at one end. c A drawing pin has a sharp point at one end and a large, flat head at the other broad head force sharp point

1 answer:

nlexa [21]3 years ago

4 0

Answer:

a cricket stamp

Explanation:

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6. A warehouse employee is pushing a 15.0 kg desk across a floor at a constant speed of 0.50 m/s. How much work must the employe

MariettaO [177]

Answer:

7.5 J

Explanation:

To answer the question given above, we need to determine the energy that will bring about the speed of 1 m/s. This can be obtained as follow:

Mass (m) = 15 Kg

Velocity (v) = 1 m/s

Energy (E) =?

E = ½mv²

E = ½ × 15 × 1²

E = ½ × 15 × 1

E = ½ × 15

E = 7.5 J

Therefore, to change the speed to 1 m/s, the employee must do a work of 7.5 J.

3 0

2 years ago

I SERIOUSLY can't do this type of questions so can someone solve it detailedly and putting with letters (there is a system you n

KatRina [158]

Answer:

4 Ohms

Explanation

(This is seriously not as hard as it looks :)

You only need two types of calculations:

replace two resistances, say, R1 and R2, connected in a series by a single one R. In this case the new R is a sum of the two: $R = R_1+R_2$
replace two resistances that are connected in parallel. In that case: $\frac{1}{R}= \frac{1}{R_1}+\frac{1}{R_2}\\\mbox{or}\\R= \frac{R_1\cdot R_2}{R_1+R_2}$

I am attaching a drawing showing the process of stepwise replacement of two resistances at a time (am using rectangles to represent a resistance). The left-most image shows the starting point, just a little bit "warped" to see it better. The two resistances (6 Ohm next to each other) are in parallel and are replaced by a single resistance (3 Ohm, see formula above) in the top middle image. Next, the two resistances (9 and 3 Ohm) are nicely in series, so they can be replaced by their sum, which is what happened going to the top right image. Finally we have two resistances in parallel and they can be replaced by a single, final, resistance as shown in the bottom right image. That (4 Ohms) is the <em>equivalent resistance</em> of the original circuit.

Using these two transformations you will be able to solve step by step any problem like this, no matter how complex.

5 0

3 years ago

A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a

Doss [256]

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$ .

Part(b): The speed of the wave is $\bf{0.5~m/s}$ .

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$ .

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$ .

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$ .

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$

5 0

3 years ago

A moving electron passes near the nucleus of a gold atom, which contains 79 protons and 118 neutrons. At a particular moment the

AURORKA [14]

Answer:

a) 22.471 x 10^-11 N

b) -22.471 x 10^-11 N

Explanation:

given the charge on an electron = 1.6 x 10^-19 C

nucleus contain 79 protons

hence, charge on nucleus =79 x 1.6 x 10^-19 C

distance r = 9 x 10^-9 m

a) therefore, force exerted by the gold nucleus on the electron = Kqe/r^2

= 9 x 10^9 x 79 x 1.6 x 10^-19 x 1.6 x 10^-19 / ( 9 x 10^-9 m)^2

= 22.471 x 10^-11 N

b) force exerted by the electron on the gold nucleus ; = -Force exerted by gold on electron ( From newton third law of motion)

= - 22.471 x 10^-11 N

4 0

3 years ago

speed of light is 3×10^8m/sec. the circumference equador is 4×10^4km. if light turns around equador for 5 min, how many tours ca

Nataliya [291]

Answer: 2250 tours

Explanation:

We have the following data:

Speed of light: $c=3(10)^{8}m/s$

Circumference of the equador: $r=4(10)^{4}km \frac{1000 m}{1 km}=4(10)^{7}m$

The time in which light turns around equator: $t=5 min \frac{60 s}{1 min}=300 s$

We need to find how many tours does light do around the equator in $t=300 s$

Let's begin by the expression of the speed, which is a relation between the traveled distance ( $d$ ) and time:

$c=\frac{d}{t}$ (1)

Isolating $d$ :

$d=c.t$ (2)

$d=(3(10)^{8}m/s)(300 s)$ (3)

$d=9(10)^{10}m$ (4) This is the distance light travels in 5 min.

Now we need to know to how many tours is this distance equivalent, we can know this by a Rule of three:

1 tour ---- $r$ (circumference of the equator)

? tour ---- $d$

Then:

$?=\frac{(1 tour)(d)}{r}$

$?=\frac{(1 tour)(9(10)^{10}m)}{4(10)^{7}m}$

$?=2250 tours$ This is the count of tours in 5 minutes.

8 0

3 years ago