All categories

2 years ago

a beaker has mass of 125 g. what is the mass of liquid if the beaker plus liquid have a mass of 232 g?

Physics

1 answer:

MrRissso [65]2 years ago

7 0

Answer: 107g

Explanation:

Just subtract 125 from 232, like this: 232-125=107g.

You might be interested in

At t = 0, a car registers at 30 miles/hr. Forty seconds later, the car’s velocity is now at 50 miles/hr. Assuming constant accel

ratelena [41]

D.

50 mph - 30 mph= 20 mph net velocity
change.
20mph/3600 seconds/hour= .00555 MPS
.0055 miles per second

40 seconds to complete the change
.0055/40= .000138

7 0

2 years ago

Which of the following best describes why understanding a watershed and its boundaries is important in designing housing develop

vagabundo [1.1K]

Answer:

Explanation:

5 0

2 years ago

To pop a balloon you stab it with a pencil. If the area of the pencil tip is .01 cm² and the pressure applied by the pencil to t

Romashka-Z-Leto [24]

Answer:

Push with force of 1N

Explanation:

I have explained in the paper.

Goodluck

3 0

3 years ago

A 5.0-kg box has an acceleration of 2.0 m/s2 when it is pulled by a horizontal force across a surface with μK = 0.50. Find the w

Maslowich

Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J

Explanation: In order to solve this problem we have to used the second Newton law given by:

∑F= m*a

F-f=m*a where f is the friction force (uk*Normal), from this we have

F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N

then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N

the net Force = (34.5-24.5)N= 10 N

Finally the work done by the net force is equal to kinetic energy change so

W=∫Force net*dr= 10 N* 0.1 m= 1J

7 0

3 years ago

Read 2 more answers

A ball is thrown vertically upwards with an initial velocity of 20.00 m/s. Neglecting air resistance, how long is the ball in th

algol [13]

Answer:

1) The greatest height attained by the ball equals 20.387 meters.

2) The time it takes for the ball to reach 15 meters approximately equals 1 second.

Explanation:

The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.

thus using third equation of kinematics we obtain the height attained as

$v^2=u^2+2as$

where

'v' is the final speed of the ball

'u' is the initial speed of the ball

'a' is the acceleration that the ball is under which in this case equals 9.81 $m/s^{2}$

's' is the distance it covers

Thus for maximum height applying the values in the equation we get

$0=20^{2}-2\times 9.81\times h\\\\\therefore h=\frac{20^{2}}{2\times 9.81}=20.387meters$

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as

$v^2=20^{2}-2\times 9.81\times 15\\\\v^{2}=105.7\\\\\therefore v=10.28m/s$

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

$v=u+at\\\\t=\frac{v-u}{a}\\\\t=\frac{10.28-20}{-9.81}=0.991seconds\approx 1second$

4 0

3 years ago