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Leni [432]
3 years ago
15

Match these items and the phrases with which they are associated.

Physics
1 answer:
amid [387]3 years ago
4 0
Amount of matter in object is mass.density is mass/volume.h2o is water.drew first picture of atom is Neil's Bohr.l* w* h is volume.basic unit of matter is atom.mixture is concrete.n=1 is inner shell.upward force of a liquid on an object is buoyancy.
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A student pushes a 2.85 kg cart causing it to accelerate at a rate of 4.9 m/s squared .What amount of force must the student hav
uysha [10]
In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.

7 0
3 years ago
To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You
drek231 [11]

Answer : The heat change of the cold water in Joules is, 1.6\times 10^3J

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

Density=\frac{Mass}{Volume}

Mass=Density\times Volume=1g/mL\times 45mL=45g

Now we have to calculate the heat change of cold water.

Formula used :

Q=m\times c\times (T_2-T_1)

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = 4.184J/g^oC

T_1 = initial temperature of cold water = 24.7^oC

T_2 = final temperature  = 33.4^oC

Now put all the given value in the above formula, we get:

Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC

Q=1638.036J=1.6\times 10^3J

Therefore, the heat change of cold water is 1.6\times 10^3J

4 0
3 years ago
A wheel of diameter 30.0 cm starts from rest and rotates with a constant angular acceleration of 3.50 rad/s2 . At the instant th
____ [38]

Answer: The radial acceleration of a point on the rim in two ways is 13.20 m/s^2

Explanation: Please see the attachments below

5 0
3 years ago
A 5.3 kg cat and a 2.5 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. how far to the left of the pivot must
nika2105 [10]
Moment about the pivot must be equal for the seesaw to balance. Initially, the first cat and the bowl are at 2 m from the pivot.

The moment due to cat = 5.3*2 = 10.6 kg.m
The moment due to bowl = 2.5*2 = 5 kg.m
The unbalanced moment = 10.6 - 5 = 5.6 kg.m

Therefore, the 3.7 kg cat should stand at a distance x from the pivot in left to balance the 5.6 kg.m.
That is,
3.7*x = 5.6 => x = 5.6/3.7 = 1.5134 m to the left (on the side of the bowl)
7 0
3 years ago
In longitudinal waves the places where the coils are bunched together are called *
Firdavs [7]
In longitudinal waves the places where the coils are bunched together are called *
Compressions
3 0
2 years ago
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