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PIT_PIT [208]
3 years ago
15

Amit has found a rock in his backyard and would like to measure its mass, volume, and density. What tools can he use to perform

these measurements? A.  meter stick, balance, and graduated cylinder B.  graduated cylinder and balance C.  meter stick and balance D.  graduated cylinder and meter stick
Physics
1 answer:
Simora [160]3 years ago
7 0
A. Should be the correct answer:) hope this helped! :)
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You start pushing a suitcase full of clothes in the horizontal direction with a force of 25.0 newtons. The weight of the suitcas
sasho [114]

Answer:

Distance traveled will be 5.6307 m

Explanation:

Time t = 3 sec

We have given force F = 25 N

We know that force is given by F = ma

So ma = 25 -----------eqn 1

Weight is given by W = 196 N

We know that weight is given by W = mg

So mg = 196 -----------------eqn 2

From equation 1 and equation 2 \frac{a}{g}=\frac{25}{196}

a=1.2512m/sec^2

Initial velocity is given as 0 so u = 0 m/sec

From second equation of motion s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 1.2512\times 3^2=5.6307m

7 0
4 years ago
Calculate the density of a tin of mass 100g whose dimensions are 2cmx5cmx​
Luda [366]

Answer:

your question in not complete.

you need to the high too.

7 0
3 years ago
Each plate of an air-filled parallel-plate capacitor has an area of 45.0 cm2, and the separation of the plates is 0.080 mm. A ba
maw [93]

Answer:

Option (e)

Explanation:

A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,

Energy density = 100 J/m

Let Q be the charge on the plates.

Energy density = 1/2 x ε0 x E^2

100 = 0.5 x 8.854 x 10^-12 x E^2

E = 4.75 x 10^6 V/m

V = E x d

V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V

C = ε0 A / d

C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F

Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C

Q =  190 nC

3 0
3 years ago
In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m
ikadub [295]

(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

p_y =0

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2

where

m is the mass of each ball

v_1= 4.60 m/s is the velocity of the cue ball after the collision

v_2 = 3.40 m/s is the velocity of the second ball after the collision

\phi_1=28.0^{\circ} is the angle of the cue ball with the x-axis

\phi_2 is the angle of the second ball

Solving for \phi_2, we find the angle between the direction of motion of the second ball and the original direction of motion:

sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

p_x = m u

where

m is the mass of the cue ball

u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

7 0
4 years ago
Consider two less-than-desirable options.
Gemiola [76]

Answer:

The force would be the same in both cases - option C.

Explanation:

The change in momentum is known as an impulse. In the two cases under consideration, the change in momentum is the same, thus impulse for both cases is the same.

Impulse is the average force multiplied by time interval.

I = F(average)*ΔT. Where F(average) is the average force and ΔT is the time interval.

The average force in both cases is the same since the collision time is the same.

Thus option C is the correct answer.

7 0
3 years ago
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