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Kisachek [45]
3 years ago
14

How many substances/elements are on the periodic table?​

Physics
1 answer:
Morgarella [4.7K]3 years ago
7 0

Answer:

There are 118 substances/elements on the periodic table.

Explanation:

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A 60kg60 kg board that is 6 m6 m long is placed at the edge of a platform, with 4 m4 m of its length extending over the edge. Th
scoundrel [369]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The minimum mass of M_1 = 90\  kg correct option is  E

Explanation:

 Free body diagram of the set up  in the question is shown on the third uploaded image

  The mass of board is  M = 60kg

   The length of the board is L =  6 \ m

    The length extending over the edge is L_e = 4 \ m

    The second mass is  M_2 = 30kg

Now to obtain M_1 we take moment about the edge of the platform

               M_1 g L_1 = Mg \frac{L}{2}  + M_2 g L_2

              M_1  L_1 = M \frac{L}{2}  + M_2  L_2

  Substituting value  

               M_1 (2) = (60)(1) + (30)(4)

               M_1  = 90 \ kg

8 0
4 years ago
A car needs to generate 75.0 hp in order to maintain a constant velocity of 27.3 m/s on a flat road. What is the magnitude of th
MaRussiya [10]

Answer:

2048.63 N

Explanation:

Force: The product of mass and the  acceleration of a body. The S.I unit of Force is Newton(N).

From the question

P = F×V.................... Equation 1

Where P = power generated by the car, F = Total resistive force acting on the car, V = Velocity of the car.

make F the subject of the equation

F = P/V................ Equation 2

Given: P = 75.0 hp = (75×745.7) = 55927.5 W, V = 27.3 m/s

Substitute into equation 2

F = 55927.5/27.3

F = 2048.63 N

Hence the total resistive force acting on the car = 2048.63 N

6 0
4 years ago
(a) The Eskimo pushes the same 50.0-kg sled over level ground with a force of 1.75 102 N exerted horizontally, moving it a dista
vivado [14]

Answer: 0.306

Explanation:

from the question we are given the following

mass of sled (m) = 50 kg

force (f) = 1.75 x 10^2 N = 175 N

distance (s) = 6 m

net work done on the sled = 1.50 x 10 ^2 N = 150 N

acceleration due to gravity (g)  = 9.8 m/s^2

coefficient of friction = μ

lets first calculate the frictional force (ff)

ff =  μ x m x g = μ  x 50 x 9.8 = 490 μ

work done on the slide by the applied force (W1)=  f x s = 175 x 6 = 1050 j

work done on the slide by frictional force (W2) = ff x s = 490 μ x 6 = 2940μ j

now the net work done is the work done by the frictional force subtracted from the work done by the applied force

net work done = W1 - W2

150 = 1050 -  2940μ

2940μ = 1050 - 150

μ = 900 / 2940

μ = 0.306

3 0
3 years ago
What advantage does a telescope in space have over one located on the ground.
klio [65]
A telescope in space can focus on any object while a telescope on the ground cannot focus on the object it is on
7 0
4 years ago
What's effect called?<br>​
k0ka [10]

Answer:

what do u mean

Explanation:

8 0
3 years ago
Read 2 more answers
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