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2 years ago

9

How many digits are in front of the decimal in scientific notation?

1 answer:

nikklg [1K]2 years ago

7 0

Answer:

1

Explanation:

How many digits are in front of the decimal in scientific notation?

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A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t

tresset_1 [31]

Answer:

Part a)

$Reading = 2.00 kg$

Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part b)

When elevator is decending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

$F_{net} = ma$

$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0

2 years ago

a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w

nataly862011 [7]

Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load distance=10 cm

We have to find the effort distance.

We know that

$load\times load\;distance=Effort\times effort\;distance$

Using the formula

$800\times 10=200\times effort\;distance$

Effort distance= $\frac{800\times 10}{200}$

Effort distance= $\frac{8000}{200}$

Effort distance=40 cm

Hence, the effort distance will be 40 cm.

7 0

2 years ago

I NEED HELP WITH THIS QUESTION ASAP PLEASE!

daser333 [38]

Wood isn’t a medium. Pls give brainliest

8 0

2 years ago

Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is

Nesterboy [21]

Answer:

r₁/r₂ = 1/2 = 0.5

Explanation:

The resistance of a wire is given by the following formula:

R = ρL/A

where,

R = Resistance of wire

ρ = resistivity of the material of wire

L = Length of wire

A = Cross-sectional area of wire = πr²

r = radius of wire

Therefore,

R = ρL/πr²

<u>FOR WIRE A</u>:

R₁ = ρ₁L₁/πr₁² -------- equation 1

<u>FOR WIRE B</u>:

R₂ = ρ₂L₂/πr₂² -------- equation 2

It is given that resistance of wire A is four times greater than the resistance of wire B.

R₁ = 4 R₂

using values from equation 1 and equation 2:

ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²

since, the material and length of both wires are same.

ρ₁ = ρ₂ = ρ

L₁ = L₂ = L

Therefore,

ρL/πr₁² = 4ρL/πr₂²

1/r₁² = 4/r₂²

r₁²/r₂² = 1/4

taking square root on both sides:

<u>r₁/r₂ = 1/2 = 0.5</u>

6 0

3 years ago

The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbi

SOVA2 [1]

Answer:

5.024 years

Explanation:

T1 = 1 year

r1 = 150 million km

r2 = 440 million km

let the period of asteroid orbit is T2.

Use Kepler's third law

T² ∝ r³

So,

$\left ( \frac{T_{2}}{T_{1}} \right )^2=\left ( \frac{r_{2}}{r_{1}} \right )^3$

$\left ( \frac{T_{2}}{1} \right )^2=\left ( \frac{440}{150} \right )^3$

T2 = 5.024 years

Thus, the period of the asteroid's orbit is 5.024 years.

4 0

2 years ago