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3 years ago

Multiply 5.036×102m by 0.078×10−1, taking into account significant figures.

Physics

1 answer:

Anika [276]3 years ago

3 0

Answer : The significant digit is 6

Explanation :

Multiply $5.036\times10^{2}m$ by $0.078\times10^{-1}m$

Now, on multiplying

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.392808 \times10^{1}\ m^{2}$

$5.036\times10^{2}m \times0.078\times10^{-1}m = 0.0392808\ m^{2}$

Now, the significant digit is 6.

Hence, this is the required solution.

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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel

Zanzabum

Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m

Answer: ymax = 10084.2m

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3 years ago

Meter #1 can measure voltage to within 0.1 volts. Meter #2 can measure voltage to within 0.01 volts.

Ann [662]

Meter #2 is more precise.

There's no information here that tells us which meter is more accurate.

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3 years ago

The Venus flytrap is known for which of these behaviors?

tensa zangetsu [6.8K]

Snapping a leaf shut around an insect, I think.

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2 years ago

Holly puts a box into the trunk of her car. Later, she drives around an unbanked curve that has a radius of 48 m. The speed of t

LiRa [457]

Answer:

The minimum coefficient of friction is 0.544

Solution:

As per the question:

Radius of the curve, R = 48 m

Speed of the car, v = 16 m/s

To calculate the minimum coefficient of static friction:

The centrifugal force on the box is in the outward direction and is given by:

$F_{c} = \frac{mv^{2}}{R}$

$f_{s} = \mu_{s}mg$

where

$\mu_{s}$ = coefficient of static friction

The net force on the box is zero, since, the box is stationary and is given by:

$F_{net} = f_{s} - F_{c}$

$0 = f_{s} - F_{c}$

$\mu_{s}mg = \frac{mv^{2}}{R}$

$\mu_{s} = \frac{v^{2}}{gR}$

$\mu_{s} = \frac{16^{2}}{9.8\times 48} = 0.544$

3 0

3 years ago

In a tank full of water, the pressure on a surface 2 meters below the water level is 1.5 kPA. What's the pressure on a surface 6

Lina20 [59]

Answer:

P = 40.7kPa

Explanation:

To find the pressure on a surface 6 meter below you use the following formula, which takes into account the heights in which pressures are measured and also the density of the fluid and the gravitational acceleration:

$P_2-P_1=-\rho g(y_2-y_1)$ (1)

P2: pressure for a height of -6 m = ?

P1: pressure for a height of -2 m = 1.5kPa = 1500 Pa

ρ: density of water = 1000kg/m^3

g: gravitational acceleration = 9.8 ms^2

y2: -6m

y1: -2m

(the height is measure from the water level, because of that, the heights are negative)

You solve the equation (1) for P1:

$P_1=P_2-\rho g(y_2-y_1)$ (2)

Next, you replace the values of all variables in equation (2):

$P_2=1500Pa-(1000kg/m^3)(9.8m/s^2)(-6-(-2))m=40700Pa\\\\P_2=40.7kPa$

hence, the pressure on a surface 6 m below the water level is 40.7kPa

5 0

3 years ago