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aleksandrvk [35]
3 years ago
11

How much pressure is applied to the

Physics
1 answer:
den301095 [7]3 years ago
4 0

Answer:

306.12N/m^2

Explanation:

Mass = 30 kg

g = 9.8m/s2

F = mg

F = 30 x 9.8

F = 294N

L = 0.98m

Since the stilts is square, then the area is L^2

A = L^2

A = 0.98^2

A = 0.9604m^2

P =?

P = F/A

P = 294/0.9604

P = 306.12N/m^2

The pressure exerted by the man is 306.12N/m^2

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What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem
8090 [49]

Answer:

\Delta L = 4.2 m

Explanation:

As per the formula of thermal expansion we know that

L = L_o(1 + \alpha\Delta T)

so here we will have

L_o = 1400 m

\alpha = 12 \times 10^{-6} per ^oC

\Delta T = 250 degree C

so here change in the length of the rod is given as

\Delta L = L - L_o

\Delta L = L_o \alpha \Delta T

\Delta L = 1400 (12 \times 10^{-6})(250)

\Delta L = 4.2 m

8 0
4 years ago
HELPPP!!!!<br> if you double the force what is the change in acceleration
jenyasd209 [6]

As we know that acceleration is directly proportional to force, therefore as the force is doubled, acceleration gets doubled too.

8 0
3 years ago
Which example involves the transformation of chemical energy directly into light energy?
zimovet [89]

burning sodium or magnesium

8 0
3 years ago
An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _
Viefleur [7K]
Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
where 
f=5 cm is the focal length
p=10 cm is the distance of the object from the mirror
q is the distance of the image from the mirror.

Rearranging the equation, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}
so, the distance of the image from the mirror is q=10 cm.

1b) The image height is given by the magnification equation:
\frac{h_i}{h_o}=- \frac{p}{q}
where h_i is the heigth of the image and h_o=1 cm is the height of the object. By rearranging the equation and using p and q, we find
h_i=-h_o  \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

3) As before, we find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

And now we can use the magnification equation to find the image height:
\frac{h_i}{h_o}=- \frac{p}{q}
Rearranging it, we find
h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm
and the negative sign means the image is inverted.
5 0
3 years ago
1. You get hit with a force of 45 N by a ball with a mass of 0.75 kg. How fast was the
dlinn [17]

Answer:

60 {ms}^{ - 2}

Explanation:

As we know,

=》Force = Mass × Acceleration

=》45 N = 0.75 × Acceleration

=》Acceleration = 45 ÷ 0.75

=》Acceleration = 60

hence, the Acceleration of the ball would be. 60 meters per second square

60m {s}^{ - 2}

6 0
3 years ago
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