. Two people are pushing a car of mass 2000 kg.

Physics

1 answer:

Viktor [21]4 years ago

5 0

Answer:

two people who are not going to be able to make it to class today because of the day and then I will be there at the house and then we can go

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Cellular phones use _____?

dangina [55]

Let's see: frequency of cellular phone waves (GSM phones) is (800-1900 MHz). If we look at the table of the electromagnetic spectrum, we can see that this range is contained within the frequencies of the microwaves, which include waves in the range 300 MHz-300 GHz.

So, summarizing, the correct answer is "microwaves".

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3 years ago

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Hecto the Mornar a fursa explain i magnituse of the force acting right angle to the moment arm

mixer [17]

<h2>~<u>Solution</u> :-</h2>

Here, the <u>moment arm</u> is defined as follows;

The magnitude of two forces, which when acting at right angle produce resultant force of VlOkg and when acting at 60° produce resultant of Vl3 kg. These forces are D. gravitational force of attraction towards the centre of the earth. A sample of metal weighs 219 gms in air, 180 gms in water, 120 gms in an <em>unknown fluid</em>.

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3 years ago

MRU(movimiento rectilineo uniforme) un movil viaja en linea recta con una velocidad media de 12metros /segundos durante 9segundo

liubo4ka [24]

Answer:

a) 141.6m

b) 8.4m/s

Explanation:

a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:

$x_1=v_1t_1=(12m/s)(9s)=108m\\\\x_2=v_2t_2=(4.8m/s)(7s)=33.6m\\\\x_T=x_1+x_2=108m+33.6m=141.6m$

hence, the total distance is 141.6m

b) the mean velocity of the total trajectory is given by:

$v_m=\frac{v_1+v_2}{2}=\frac{12m/s+4.8m/s}{2}=8.4\frac{m}{s}$

hence, the mean velocity is 8.4 m/s

8 0

3 years ago

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road