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4 years ago

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Remove the ice cube tray and the plastic bag from the freezer and let them sit for 3 minutes. Then gently remove the piece of ic

e from the tray and the piece of ice from the plastic bag. Place each piece of ice on a piece of foil. Check the pieces of ice about every three minutes until you can clearly see whether there’s a difference in how fast they are melting. What do you observe? HELPPP HURRYYYY

1 answer:

Salsk061 [2.6K]4 years ago

8 0

Answer:

That when you leave the ice setting without foil, it melts slower. When you put the piece of ice in the foil it melts faster than just letting it set without foil. I really hope this helps.

Explanation:

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Block M = 7.50 kg is initially moving up the incline and is increasing speed with a = [09]____________________ m/s2 . The applie

Aliun [14]

Answer:

a) 73.2N

b) 66.6N

c) 20.28N

Explanation:

F= mg=7.5×9.8=73.5N

Force parallel to the plane Fp= 73.5sin25= 31.06N

b) Fv= 73.5 cos25= 66.6N

c) Ff= u×Fv= 0.312×66.6=20.28N

a) Normal ForceFn= F/(cos25) - Fp - Ff = ma

1.1F -31.06-20.28=7.5×3.82

1.1F -51.84=28.65

1.1F= 28.65+51.84

F= 80.49/1.1

F= 73.2N

4 0

3 years ago

PLEASE HELP MEE THIS IS DUE IN 45 MINS

guajiro [1.7K]

Answer:

The distance travelled does not depend on the mass of the vehicle. Therefore, $s = d$

Explanation:

This deceleration situation can be analyzed by means of Work-Energy Theorem, where change in translational kinetic energy is equal to the work done by friction:

$\frac{1}{2}\cdot m\cdot v^{2}-\mu\cdot m\cdot g \cdot s = 0$ (1)

Where:

$m$ - Mass of the car, in kilogram.

$v$ - Initial velocity, in meters per second.

$\mu$ - Coefficient of friction, no unit.

$s$ - Travelled distance, in meters.

Then we derive an expression for the distance travelled by the vehicle:

$\frac{1}{2}\cdot v^{2} = \mu \cdot g \cdot s$

$s = \frac{v^{2}}{\mu\cdot g}$

As we notice, the distance travelled does not depend on the mass of the vehicle. Therefore, $s = d$

3 0

3 years ago

A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

8 0

3 years ago

A 1.00 kg solid, uniform disk rolls without slipping across a level surface, translating at 3.00 m/s. If the disk's radius is 0.

cluponka [151]

Explanation:

Given that,

Mass of the disk, m = 1 kg

Translational speed of the disk, v = 3 m/s

Radius of the disk, r = 0.39 m

(a) Let K is the tranlational kinetic energy of the disk. Its formula is given by :

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}\times 1\times 3^2$

K = 4.5 J

(b) Let K' is the rotational kinetic energy of the disk. Its formula is given by :

$K'=\dfrac{1}{2}I\omega^2$

$K'=\dfrac{1}{2}\dfrac{mr^2}{2}(\dfrac{v}{r})^2$

$K'=\dfrac{1}{4}mv^2$

$K'=\dfrac{1}{4}\times 1\times 3^2$

K' = 2.25 J

Hence, this is the required solution.

5 0

3 years ago

A chemical reaction is taking place in a beaker. The side of the beaker feels cold to the touch. Which type of reaction is takin

Mekhanik [1.2K]

A chemical reaction is taking place in a beaker. The side of the beaker feels cold to the touch. An <span>endothermic reaction is taking place, becaue energy is being absorbed from the surrounding.</span>

5 0

3 years ago

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