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castortr0y [4]
2 years ago
15

Different between : Metalloids and alloys ​

Physics
2 answers:
FromTheMoon [43]2 years ago
8 0

Answer:

metalloids are chemical elements that form substances that have properties of metals and non metals while alloys is a mixture of metals

iVinArrow [24]2 years ago
7 0

Answer:

Alloys are just mixtures of metal and metalloids are metallic substances with properties of metal.

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A hockey ball accelerates from 0.m/s to 25m/s in 0.05 seconds what is the acceleration of the ball ?
S_A_V [24]

Answer:

500

Explanation:

25/0.05 .

I think this ?

3 0
3 years ago
What is the charge on 1.0 kg of protons? (e = 1.60 × 10-19 c, mproton = 1.67 × 10-27 kg)?
Andru [333]
First, we need to find the number of protons, which is the total mass divided by the mass of one proton:
N= \frac{m}{m_p}= \frac{1.0 kg}{1.67 \cdot 10^{-27} kg}=6.0 \cdot 10^{26} protons

Then, the total charge is the number of protons times the charge of a  single proton:
Q=Ne = 6.0 \cdot 10^{26}\cdot 1.60 \cdot 10^{-19} C=9.6 \cdot 10^7 C
8 0
3 years ago
Read 2 more answers
Jazz is a 172 lb athlete who exercised at 7.6 METs. At this workload, what is his energy expenditure in kcals/min.? Round to the
dezoksy [38]

Answer:

Energy expenditure in K cals/min = 10 K cals /min (approximately)

Explanation:

As we know

Energy expenditure in Kcal/min=  METs x 3.5 x Body weight (kg) / 200

Given is METs=7.6

Weight of Jazz= 172lb=78.02kg

putting the values in formula,

Energy expenditure in K cals/min=  7.6 x 3.5 x 78.02 / 200

                                                       =10.38 K cals /min

                                                       =10 K cals /min (approximately)

Therefore, Energy expenditure in K cals/min by Jazz will be approximately 10 K cals /min

8 0
3 years ago
A rod 7.0 m long is pivoted at a point 2.0 m from the left end. A downward force of 50 N acts at the left end, and a downward fo
kicyunya [14]

If the rod is in rotational equilibrium, then the net torques acting on it is zero:

∑ τ = 0

Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:

• at the left end,

τ = + (50 N) (2.0 m) = 100 N•m

• at the right end,

τ = - (200 N) (5.0 m) = - 1000 N•m

• at a point a distance d to the right of the pivot point,

τ = + (300 N) d

Then

∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0

⇒   (300 N) d = 1100 N•m

⇒   d ≈ 3.7 m

6 0
2 years ago
How much force is required to accelerate a 12 kg mass at 5 m/s 2
Savatey [412]

Answer:

60 N

Explanation:

This is just Newton's Second Law

F = m*a

F = ?

m = 12 kg

a = 5 m/^2

F = 5*12 = 60 Newtons

4 0
2 years ago
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