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I am Lyosha [343]
3 years ago
11

A projectile is fired horizontally from a height of 78.4 m at a speed of 300 m/sec. How far did it travel horizontally before hi

tting the ground?​
Physics
1 answer:
Mice21 [21]3 years ago
3 0

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g

U is the speed = 300m/s

H is the maximum height = 78.4m

g is the acceleration due to gravity = 9.8m/s²

Substitute into the fromula;

R = 300√2(78.4)/9.8

R = 300 √(16)

R = 300*4

R = 1200m

Hence the projectile travelled 1200m before hitting the ground

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Answer:

Low-temperature blackbody

Explanation:

There are 3 types of blackbody temperatures.

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A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.

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a person starts at a position of 2 meters and finishes at a postion of 25 meters. the trip takes 4.5 seconds. what is the person
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Explanation:

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Calculate the size of the image of a tree that is 8m high and 80 m a pinhole camera that is 20 cm long . what is its magnificati
Vladimir79 [104]

1) Size of the image: 2 cm

In order to calculate the size of the image, we can use the following proportion:

p:q = h_o : h_i

where

p = 80 m is the distance of the tree from the pinhole

q = 20 cm = 0.2 m is the distance of the image from the pinhole

h_o = 8 m is the heigth of the object

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By re-arranging the proportion, we find

h_i = \frac{h_o \cdot q}{p}=\frac{(8 m)(0.2 m)}{80 m}=0.02 m=2 cm


2) Magnification: 0.0025

The magnification of a camera is given by the ratio between the size of the image and the size of the real object:

M=\frac{h_i}{h_o}

so, in this problem we have

M=\frac{0.02 m}{8 m}=0.0025


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I hope I helped:)
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