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3 years ago

An electric grinder uses a grinding wheel

Physics

2 answers:

Leona [35]3 years ago

7 0

An electric grinder uses a grinding wheel of 12-cm radius.

luda_lava [24]3 years ago

3 0

(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s²

(B) 
(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev 

(C) 
(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s

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4. A group of students made a rocket and launched it vertically upwards with velocity

Aleks04 [339]

Answer:

Approximately $72.9\; \rm m$ , assuming that the rocket had no propulsion onboard, and that air resistance on the rocket is negligible.

Explanation:

Initial velocity of this rocket: $u = 27\; \rm m\cdot s^{-1}$ .

When the rocket is at its maximum height, the velocity of the rocket would be equal to $0$ . That is: $v = 0\; \rm m \cdot s^{-1}$ .

The acceleration of the rocket (because of gravity) is constantly downwards, with a value of $a = -g = -10\; \rm m \cdot s^{-2}$ .

Let $x$ denote the distance that the rocket travelled from the launch site to the place where it attained maximum height. The following equation would relate $x \!$ to $u$ , $v$ , and $a$ :

$\displaystyle x = \frac{v^2 - u^2}{2\, a}$ .

Apply this equation to find the value of $x$ :

$\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{\left(0\; \rm m\cdot s^{-1}\right)^{2} - \left(27\; \rm m \cdot s^{-1}\right)^{2}}{2 \times 10\; \rm m \cdot s^{-2}} = 36.45\; \rm m\end{aligned}$ .

In other words, the maximum height that this rocket attained would be $36.45\; \rm m$ .

Again, assume that the air resistance on this rocket is negligible. The rocket would return to the ground along the same path, and would cover a total distance of $2\times 36.45\; \rm m = 72.9\; \rm m$ .

7 0

3 years ago

Need help before 10pm tomorrow night

konstantin123 [22]

The first one is A and the second one would be C

5 0

3 years ago

A Navy vessel is traveling due north during wartime. A torpedo has been launched by an enemy directly toward the stern (rear) of

Anika [276]

Answer:

The correct option is;

B) No, the Navy vessel is slower

Explanation:

The speed of some torpedoes can be as high as 370 km/h. The average speed of a fast Navy vessel is approximately 110 km/h

Therefore, the torpedoes travel approximately 3 times as fast as the (slower) Navy vessel, such that the torpedo covers three times the distance of the Navy vessel in the same time and therefore, if the Navy vessel and the torpedo continue in a straight line (in the same direction) due north the vessel can not outrun the torpedo

Therefore, no the Navy vessel travels slower than a torpedo.

8 0

3 years ago

A penny rolls along the table for a distance of 1.3 meters. Jackie pushes it 40 centimeters further in the same direction.

brilliants [131]

40 meters times 1 meter over 100 centimeters equals 0.4 meters. 1.3 meters + 40 centimeters =. 1.3 m + 0.4 m = 1.7 m. The answer is 1.7 meters

7 0

4 years ago

PRACTICE ANOTHER A cube of wood having an edge dimension of 19.7 cm and a density of 647 kg/m3 floats on water. (a) What is the

dedylja [7]

CHECK COMPLETE QUESTION

The cube of wood having an edge dimension of 19.7cm and a density of 647kg/m3 floats on water

(a) What is the distance from the horizontal top surface of the cube to the water level answer in cm

(b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface answer in kg

Answer:

a)6.29cm

b)2.78 kg

Explanation:

Given:

Let us calculate the volume first, we were given dimension as 19.7cm=0.197m

Volume is (0.197 meters)³ = 0.00764m³

Then we can calculate the mass as;

Given mass is 647 kg/m³ x 0.00774m³ = 4.947kg

The weight = mass × acceleration due to gravity

weight = 4.947 x 9.8 N/kg = 48.44N

By Floating we can say the the buoyancy force has to equal the weight (48.44 N) which has

which is equal to the weight of volume of the displaced water. Or the mass, the calculation is the same.

We know that density of fresh water at 20ºC is 998 kg/m³

Then we can calculate the volume of displaced water as

4.947 kg / 998 kg/m³ = 0.00496 m³

We know that the displaced water has a shape of a rectangular solid with 0.197 meters on the two horizontal dimensions, and h as the height to the surface then

V = 0.197²h = 0.00496

0.00496= 0.197²h

h = 0.1278 meters or 12.78 cm

Then the the distance exposed, would be 19.7–12.78 = 6.29 cm

b) ifthe cube is fully submerged, the volume of the displaced water is 0.00774m³

mass of displaced water is 0.00774m³ x 998 kg/m³ = 7.724 kg

Added mass is the mass of the displaced water – mass of block

= 7.724–4.947 = 2.78 kg

5 0

3 years ago