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N76 [4]
2 years ago
9

Which of the following scenarios represents a safe situation for the child?

Physics
2 answers:
butalik [34]2 years ago
8 0

Answer:

B---its B but I need 20 characters so jfhhfhdhfhfhfhdbdbdhdf fhfhfbfb fjf f bbn HD fjf

alisha [4.7K]2 years ago
5 0

Answer:

the answer is b

Explanation:

cause I said

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Subtract the mass of the cylinder from the mass of the cylinder when it contains gasoline. This is the mass of the gasoline. Divide this figure by the volume, 100 ml, to get the density.

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2 years ago
How is work calculated when the force applied is not parallel to the displacement
Leno4ka [110]
Pretty sure it’s Force*Distance*Cos(theta)
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The value of 1.0004 to the power 1 by 2 using Binomial approximation is​
IgorLugansk [536]

Given:

The given value is (1.0004)^{\frac{1}{2}}.

To find:

The value of the given expression by using the Binomial approximation.

Explanation:

We have,

(1.0004)^{\frac{1}{2}}

It can be written as:

(1.0004)^{\frac{1}{2}}=(1+0.0004)^{\frac{1}{2}}

(1.0004)^{\frac{1}{2}}=1+\dfrac{1}{2}\times 0.0004      [\because (1+x)^n=1+nx]

(1.0004)^{\frac{1}{2}}=1+0.0002

(1.0004)^{\frac{1}{2}}=1.0002

Therefore, the approximate value of the given expression is 1.0002.

3 0
2 years ago
You accidentally drop an eraser out of the window of an apartment 15 m above the ground
docker41 [41]

Answer:

hello, yes or nou sorry jaja

3 0
2 years ago
A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th
zhannawk [14.2K]

Answer: V_{f}=2.96m/s    

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight w of the block has an x-component and y-component:

w_{x}=wsin(\theta)    (1)

w_{y}=wcos(\theta)    (2)

As well as the Normal Force N:

N_{x}=Nsin(\theta)    (3)

N_{y}=Ncos(\theta)    (4)

In addition, we know N=w, then \sum F_{y}=0

In the X-component:

\sum F_{x}=m.a

m.a=w_{x}    (5)

Substituting (1) in (5):

wsin(\theta)=m.a    (6)

In addition, we know w=m.g, where m is the mass of the block and g the gravity acceleration, which is equal to 9.8m/{s}^{2}  

So:

m.g.sin(\theta)=m.a   (7)

a=g.sin(\theta)    (8)

a=5.88m/{s}^{2}    (9)   >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance d with the acceleration a and time t:

d=\frac{1}{2}a{t}^{2}   (10)

We already know the value of  d and calculated a, we have to find t:

t=\sqrt{\frac{2d}{a}}   (11)

t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}   (12)

t=0.50s   (13) >>>This is the time it takes to the block to go from the initial velocity V_{o} to its final velocity V_{f}

If the acceleration is the variation of the velocity in time, we can use the following equation to find V_{f}:

V_{f}-V_{o}=a.t   (13)

If V_{o}=0

V_{f}=a.t   (14)

V_{f}=(5.88m/{s}^{2})(0.50s)   (15)

Finally we get the value of the Final Velocity of the block:

V_{f}=2.96m/s    

6 0
3 years ago
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