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3 years ago

Suppose the value of one division of vernier scale is 0.5mm and the value of one main scale

Physics

1 answer:

gregori [183]3 years ago

8 0

Answer:

-0.01 mm

Explanation:

We are given that

The value of one division of vernier scale =0.5 mm

The value of one main scale division=0.49 mm

We have to find the value of least count of the instrument in mm.

We know that

Leas count of vernier caliper=1 main scale division-1 vernier scale division

Least count of vernier caliper=0.49-0.50=-0.01 mm

Hence, the least count of the instrument=-0.01 mm

Answer: -0.01 mm

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Wildlife biologists are attempting to monitor the population size of the insect pest known as the gypsy moth (Lymantria dispar)

Elden [556K]

Answer: Option A: The number of trees sampled.

The accuracy can be understood as how close is the measured value to the true value. The aim is to monitor the population size of the insect pest in a 50 square kilometer. Random trees are selected, and number of eggs and larvae are counted. So, the measured value would be closer to actual value when the number of trees sampled are increased. More the number of trees sampled, less would be the chances of error and the accuracy of the estimate would increase.

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3 years ago

Read 2 more answers

Convert 1 mm to meters using scientific notation. Include units in your answer.

miskamm [114]

Answer:

1.0 x 10^-3 m

Explanation:

1mm = (1/1000) m = 0.001m = 1.0 x 10^-3m (because to get 1.0, you have to movi the decimal point in the "0.001" three spots over to the right)

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2 years ago

What 2 things need to be known about an object in order to determine its kinetic energy?

NikAS [45]

I believe the answer is the mass of the object and the speed at which it is moving.

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3 years ago

Ms. Sparkle bought 12 cans of diet soda. Each can

Anit [1.1K]

4.2 liters..... there are 1,000 mL in a liter and there is a total of 4200 mL in this case which is divided by 1000 which gives you 4.2 liters.

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2 years ago

When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, th

Fofino [41]

Answer:

Option E is correct 310N

Explanation:

Given that the force used to push the crate is F = 200N

The force directed 20° below the horizontal

Mass of crate is m = 25kg

Weight of the crate can be determine using

W = mg

g is gravitational constant =9.8m/s²

W = 25×9.8

W = 245 N

Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

ΣFy = m•ay

ay = 0, since the body is not moving in the vertical or y direction

N—W—F•Sin20 = 0

N = W+F•Sin20

N = 245+ 200Sin20

N = 245 + 68.4

N = 313.4 N

The normal force is approximately 310 N to the nearest ten

3 0

2 years ago