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zhenek [66]
3 years ago
13

Match the block to its correct density. 1. Block A 0.64 Kg/L 2. Block B 0.917 kg/L 3. Block C 0.70 Kg/L 19.3 kg/L 4. Block D 3.5

3 Kg/L 5. Block E​
Physics
1 answer:
Sedbober [7]3 years ago
4 0

Answer:

quit school aghahahha

Explanation:

xD

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I am Lyosha [343]
Life darling
☆*:.。. o(≧▽≦)o .。.:*☆
7 0
3 years ago
Read 2 more answers
Consider a simple pendulum with a period
lisabon 2012 [21]

Answer:

1. The length is 8.35m

2. The period on the moon is 14.05 secs

Explanation:

1. Data obtained from the question. This includes the following:

Period (T) = 5.8 secs

Acceleration due to gravity (g) = 9.8 m/s2

Length (L) =...?

The length can be obtained by using the formula given below:

T = 2π√(L/g)

5.8 = 2π√(L/9.8)

Take the square of both side

(5.8)^2 = 4π^2 x L/ 9.8

Cross multiply

4π^2 x L = (5.8)^2 x 9.8

Divide both side by 4π^2

L = (5.8)^2 x 9.8 / 4π^2

L= 8.35 m

2. Data obtained from the question. This includes the following:

Acceleration due to gravity (g) = 1.67 m/s2

Length (L) = 8.35m (the length remains the same)

Period (T) =?

The period can be obtained as follow:

T = 2π√(L/g)

T = 2π√(8.35/1.67)

T = 14.05 secs

Therefore, the period on the moon is 14.05 secs

4 0
3 years ago
What is the average salt content of seawater?
marta [7]
This gots to be the answer, average, seawater in the world's oceans has a salinity of approximately 3.5%, or 35 parts per thousand.
5 0
3 years ago
A 6.3 g bullet leaves the muzzle of a rifle with a speed of 596.2 m/s. what constant force is exerted on the bullet while it is
ivanzaharov [21]
<span>anwser will be 

F = ma

where 

F = force exerted on the bullet 
m = mass of the bullet = 5 gm (given) = 0.005 kg. 
a = acceleration of the bullet 

Substituting appropriately, 

F = 0.005a --- call this Equation 1 

Next working equation is 

Vf^2 - Vo^2 = 2as 

where 

Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given) 
Vo = initial velocity of bullet = 0 
a = acceleration of bullet 
s = length of the rifle's barrel 

Substituting appropriately, 

326^2 - 0 = 2(a)(0.83) 

a = 64,022 m/sec^2 

the anwser will be
Substituting this into Equation 1, 

F = 0.005(64,022) 

F =320.11 Newtons 

Hope this helps. </span><span>
</span>
8 0
3 years ago
A wagon with an initial velocity of 2 m/s and a mass of 60 kg, gets a push with 150 joules of
Aleks04 [339]

Answer:

v_f = 3 m/s

Explanation:

From work energy theorem;

W = K_f - K_i

Where;

K_f is final kinetic energy

K_i is initial kinetic energy

W is work done

K_f = ½mv_f²

K_i = ½mv_i²

Where v_f and v_i are final and initial velocities respectively

Thus;

W = ½mv_f² - ½mv_i²

We are given;

W = 150 J

m = 60 kg

v_i = 2 m/s

Thus;

150 = ½×60(v_f² - 2²)

150 = 30(v_f² - 4)

(v_f² - 4) = 150/30

(v_f² - 4) = 5

v_f² = 5 + 4

v_f² = 9

v_f = √9

v_f = 3 m/s

7 0
2 years ago
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