All categories

3 years ago

Hola me pueden ayudar con este problema de física....por favor mirar la imagen.. gracias

Physics

1 answer:

makkiz [27]3 years ago

8 0

Si la velocidad es 3 m/s, y ellos quieren saber la distancia despues 2 segundos, necesita que multiplicar 2 y 3.

La respeusta debiera ser 6m.

You might be interested in

Which phase of matter does not conduct heat well?

inessss [21]

Gases do not conduct heat well.

4 0

3 years ago

Read 2 more answers

The mass of a truck with its cargo is 1800 kg. The truck is coasting to the right with a speed of 10 m/s. As it coasts its cargo

OverLord2011 [107]

Here we apply conservation of linear momentum. The momentum of the truck with cargo and without cargo remains constant. That is,

$mv=m'v'$ .

Here $m,v$ are initial mass and velocity. $m',v'$ are final mass and velocity. Here $m=1800,v=10$ and $m'=1800-300=1500$ .

The velocity of the truck be after its cargo is taken off is

$v'=\frac{mv}{m'} \\ v'=\frac{1800*10}{1500} \\ v'=12 m/s$

6 0

4 years ago

Read 2 more answers

What is the resolution of an analog-to-digital converter with a word length of 12 bits and an analogue signal input range of 100

Elena L [17]

Answer:

The resolution of an analog-to-digital converter is 24.41 mV

Explanation:

Resolution of an analog-to-digital = (analogue signal input range)/2ⁿ

where;

n is the number or length of bit, and in this question it is given as 12

Also, the analogue signal input range is 100V

Resolution of an analog-to-digital = 100V/2¹²

2¹² = 4096

Resolution of an analog-to-digital = 100V/4096

Resolution of an analog-to-digital = 0.02441 V = 24.41 mV

Therefore, the resolution of an analog-to-digital converter is 24.41 mV

5 0

4 years ago

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago

A floating boat has a weight of 100 N. What is the buoyant force on the boat?

lara31 [8.8K]

If the boat is floating, then it's just sitting there, and not accelerating
up or down. That means the vertical forces on it must be balanced.

So if its weight (acting downward) is 100 newtons, then the buoyant
force on it (acting upward) must also be 100 newtons.

4 0

3 years ago