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iren2701 [21]
3 years ago
7

In diving to a depth of 248 m, an elephant seal also moves 296 m due east of his starting point. What is the magnitude of the se

al's displacement?
Physics
1 answer:
Vinvika [58]3 years ago
4 0

Answer:

The displacement is 386.16m

Explanation:

A seal dives to a depth of 248m. To find displacement, we must calculate the resultant vectors which will give us the displacement

R= sqrt(vector1+vector2)

Since this is a right angle triangle

R= sqrt(248^2 + 296^2)

R= sqrt(149120)

R= 386.16m

Displacement = 386.16m

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The magnification of a microscope is increased when_________.
azamat

Answer:

Option B

Explanation:

Magnification of Microscope is  

M = M_o \times M_e  

Mo= Magnification of objective lens and  

Me= magnification of the eyepiece.  

Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.  

Magnification,  

M \propto \dfrac{1}{f}

when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.

Thus. Magnification will increase by decreasing the focal length.

The correct answer is Option B

6 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
3 years ago
PLZ HELP I WILL GIVE BRAINLIEST
Vaselesa [24]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of diver  = 77kg

Height of jump  = 8.18m

Unknown:

Final velocity  = ?

Solution:

To solve this problem, we apply the motion equation below:

             v²   = u²  + 2gH

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

H is the height

 Now insert the parameters and solve;

       v² = 0²  +  2 x 9.8 x 8.18

     v  = 12.7m/s

8 0
2 years ago
What is the atomic weight of a penny?
Gnom [1K]
I actually know the answer to this one, you use pennies to find the atomic weight of a penny, it really doesn't have a weight. LOL

4 0
3 years ago
Read 2 more answers
What is the normal force in the image below?
saveliy_v [14]

Answer:

980.7N

Explanation:

F = m * g

F = 100 * 9.807m/s^{2}

F = 980.7N

3 0
2 years ago
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