Can u anser 5,6 on the picture

A 500g cart moving at 0.25 m/s collides and sticks to a stationary 750g cart. How fast do the two carts

Tresset [83]

Answer:

0.1 m/s

Explanation:

Please see attached photo for explanation.

Mass of 1st cart (m₁) = 500 g

Initial velocity of 1st cart (u₁) = 0.25 m/s

Mass of 2nd cart (m₂) = 750 g

Initial velocity of 2nd cart (u₂) = 0 m/s

Velocity (v) after collision =.?

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(500 × 0.25) + (750 × 0) = v(500 + 750)

125 + 0 = v(1250)

125 = 1250v

Divide both side by 1250

v = 125 / 1250

v = 0.1 m/s

Thus, the two cart will move with a velocity of 0.1 m/s after collision.

3 0

3 years ago

A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di

Oksi-84 [34.3K]

Explanation:

The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.

The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.

Total displacement = $135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East$

Total time travelled = $1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr$

$\text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}$

∴ Driver's average velocity = $82.353\text{ }\frac{km}{hr}$

3 0

3 years ago

What is the vertical displacement below any line?

Montano1993 [528]

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4 0

3 years ago

Calculate the molarity of a 10. 0% cacl₂ solution. The density of the solution is 1. 0835 g/cm³.

brilliants [131]

The molarity of 10% CaCl2 is 0.9%

concentration of the given salt CaCl₂ = 10%

Density of a solution = 1.0835 g/cm³

Volume = m / d

= 100 / 1.0835

= 92.29 litres

Density = mass / volume

1.0835 × 92.29 = mass

mass = 99.99 gram

Thus the molarity can be calculated by = moles of solute / volume of solution multiplied by 100

= 0.9008/ 92.29 X 100 %

= 0.009 X 100 %

= 0.9 %

The molarity of 10% CaCl2 is 0.9%

To know more about density and molarity you may visit the link which is mentioned below:

brainly.com/question/10710093

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5 0

1 year ago

(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind

Vilka [71]

There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
$F_f = \mu N$
where $\mu$ is the coefficient of friction, while N is the normal force. So we have:
$F=\mu N$
since we know that F=300 N and $\mu=0.3$ , we can find N, the magnitude of the normal force:
$N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N$

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is $\mu=0.03$ due to the presence of the oil. Therefore, we have:
$N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N$

8 0

3 years ago