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Nutka1998 [239]

3 years ago

9

El carrito a control remoto de Juanito corre en una pista recta de 40 m de un extremo a otro en 90 segundos; luego de regreso so

bre la pista recorre 15 metros y se ubica en medio en 25 segundos, cual es la rapidez?
y cual es la velocidad?

1 answer:

PSYCHO15rus [73]3 years ago

8 0

Explanation:

hir ceh3co8d ueirfkgrewkfgv ruk gr

You might be interested in

Friends Burt and Ernie stand at opposite ends of a uniform log that is floating in a lake. The log is 3.0 m long and has mass 20

Taya2010 [7]

Answer:

The distance the log has moved by the time Ernie reaches Bur is 1.33 m.

Explanation:

give information:

The log is 3.0 m long and has mass 20.0 kg.

Burt has mass 30.0 kg; Ernie has mass 40.0 kg

Ernie has mass 40.0 kg.

to find the distance, first, we have to calculate the center of mass

X = ∑ m x /∑m

= (20 x (3/2)) + (30 x 0) + (40 x 3)/ (20+30+40)

= 150/90

= 5/3

when Ernie walk, the center of the mass is

X = (70 x 0) + (20 x (3/2))/(70 + 20)

= 30/90

= 1/3

the distance of log moved = 5/3 - 4/3 = 1.33 m

5 0

3 years ago

What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2

Margarita [4]

Answer:

$m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )$

Explanation:

We have to make a hollow sphere of inner radius $r_1$ and outer radius $r_2$ .

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

$V_t=\frac{4}{3} \pi.r_2^3$

And the volume of the hollow space in the sphere:

$V_h=\frac{4}{3} \pi.r_1^3$

Therefore the net volume of material required to make the sphere:

$V=V_t-V_h$

$V=\frac{4}{3} \pi(r_2^3-r_1^3)$

Now let the density of the of the material be $\rho$ .

<u>Then the mass of the material used is:</u>

$m=\rho.V$

$m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )$

4 0

3 years ago

1. In what type of ecosystem can monkeys population be found?

sp2606 [1]

I, thank you for posting your question here at Brainly.

Here are the answers to your questions:

1. Monkeys live in tropical rainforest ecosystems.
2. Animals should be left alone in nature to take part in the balance of the ecosystem, and not upset it. A sudden increase of monkey population, probably due to destruction of rainforests which lead to lesser area of living for them.
3. Monkeys take a huge part in seed dispersal among flesh fruit plants. Without them, it will greatly affect the conservation of the forests.

3 0

3 years ago

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

5 0

3 years ago

A 2.3 kg particle-like object moves in a plane with velocity components vx = 40 m/s and vy = 75 m/s as it passes through the poi

Sonbull [250]

Answer:

(a) $\overrightarrow{L}=885.5\widehat{k}$

(b) $\overrightarrow{L}=1046.5\widehat{k}$

Explanation:

mass, m = 2.3 kg

vx = 40 m/s

vy = 75 m/s

(a) Angular momentum is given by

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

Where, p is the linear momentum and r is the position vector about which the angular momentum is calculated.

Here, $\overrightarrow{r}=3\widehat{i}-4\widehat{j}$

$\overrightarrow{p}=m\overrightarrow{v}$

$\overrightarrow{p}=2.3\left ( 40\widehat{i}+75\widehat{j} \right )$

$\overrightarrow{p}= 92\widehat{i}+172.5\widehat{j}$

So, the angular momentum

$\overrightarrow{L}=\left ( 3\widehat{i}-4\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )$

$\overrightarrow{L}=885.5\widehat{k}$

(b) Here, $\overrightarrow{r}=(3+2)\widehat{i}+(-4+2)\widehat{j}$

$\overrightarrow{r}=5\widehat{i}-2\widehat{j}$

$\overrightarrow{L}=\left ( 5\widehat{i}-2\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )$

$\overrightarrow{L}=1046.5\widehat{k}$

6 0

3 years ago