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3 years ago

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Physics

2 answers:

Marta_Voda [28]3 years ago

3 0

Answer:

7200 m

Explanation:

The pond is 4000 meters away. The woman runs at 2.5 m/s. The time it takes her to reach the pond is:

t = 4000 m / 2.5 m/s

t = 1600 s

The dog is running at 4.5 m/s this entire time. So the distance he runs is:

d = 1600 s × 4.5 m/s

d = 7200 m

olga nikolaevna [1]3 years ago

3 0

Answer:

7200 m

Explanation:

A woman takes her pet dog on a walk to a pond which is 4km away. The woman runs with a speed of 2.5 ms-1 while the dog runs with the speed of 4.5ms^-1 towards the pond and returns to the women and then again runs back to the pond continuously until the woman also manages to reach the pond. The total distance traveled by the dog is 7200 m.

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How far can you ride your bike if you ride for 10 min at a speed of 150 m/min

Contact [7]

Answer: 1500metres

Explanation: the formula for distance is speed × time which equals 150m/min × 10 min = 1500m

This formula can be rearranged...

Speed = Distance ÷ time.

Time= Distance ÷ speed.

Distance= Time × speed.

4 0

3 years ago

Read 2 more answers

The current through a certain heater wire is found to be fairly independent of its temperature. If the current through the heate

bearhunter [10]

Answer:

Explanation:

the formular for the relationship between energy and current is E= I²Rt (Heat energy)

the heater is fairly independent of its temperature , and given the time to be constant.

with the current being doubled,

E = I² Rt

E/ I² = K

$\frac{E1}{I1^{2} }$ = $\frac{E2}{I2^{2} }$

E2 = $\frac{E1* I1^{2} }{I1^{2} }$

I2 = 2* I1

E2 = E1 * 4*I1² / I1²

dividing , we have that E2 = 4 * E1

which means during the same time interval the amount of energy will increase by a factor of four.

7 0

4 years ago

A big olive (* - 0.50 kg) lies at the origin of an xy coordinate system, and a big BrazlI nut (M - 1.5^kg) lie^s at the point (1

Afina-wow [57]

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ .

<h3>Procedure - Estimation of the displacement of the center of mass of the olive</h3>

In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> ( $\vec r$ ) and <em>static</em> conditions ( $\vec r_{o}$ ) to estimate the displacement of the center of mass of the olive ( $\overrightarrow{\Delta r}$ ):

$\vec r - \vec r_{o} = \left[\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot(m_{i}\cdot g + F_{i, x})}{\Sigma \limits_{i =1}^{2}(F_{i,x}+m_{i}\cdot g)} ,\frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot(m_{i}\cdot g + F_{i, y})}{\Sigma \limits_{i =1}^{2}(F_{i,y}+m_{i}\cdot g)} \right]-\left(\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}, \frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}\right)$ (1)

Where:

$r_{i, x}$ - x-Coordinate of the i-th element of the system, in meters.
$r_{i,y}$ - y-Coordinate of the i-th element of the system, in meters.
$F_{i,x}$ - x-Component of the net force applied on the i-th element, in newtons.
$F_{i,y}$ - y-Component of the net force applied on the i-th element, in newtons.
$m_{i}$ - Mass of the i-th element, in kilograms.
$g$ - Gravitational acceleration, in meters per square second.

If we know that $\vec r_{1} = (0, 0)\,[m]$ , $\vec r_{2} = (1, 2)\,[m]$ , $\vec F_{1} = (0, 3)\,[N]$ , $\vec F_{2} = (-3, -2)\,[N]$ , $m_{1} = 0.50\,kg$ , $m_{2} = 1.50\,kg$ and $g = 9.807\,\frac{kg}{s^{2}}$ , then the displacement of the center of mass of the olive is:

<h3>Dynamic condition $\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)} \right]$ $\vec r = (0,704, 1.233)\,[m]$ </h3>

<h3>Static condition</h3><h3> $\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)} \right]$ </h3><h3> $\vec r_{o} = \left(0.75, 1.50)\,[m]$ </h3><h3 /><h3>Displacement of the center of mass of the olive</h3>

$\overrightarrow{\Delta r} = \vec r - \vec r_{o}$

$\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]$

$\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]$

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ . $\blacksquare$

To learn more on center of mass, we kindly invite to check this verified question: brainly.com/question/8662931

3 0

2 years ago

The foot pedal on hydraulic brake system exerts a force of 45 lb on a piston with a diameter of 1.00 cm. The brake fluid in the

ElenaW [278]

Answer:

the correct answer is C, 281 lb

Explanation:

This is an exercise in fluid mechanics specifically on Pascal's principle, which states that the pressure in a fluid is equal to all points that are at the same depth

P = $\frac{F_{1} }{A_{1} }$ = $\frac{F_{2} }{A_{2} }$