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2 years ago

Can someone plz help me?

Physics

1 answer:

Luba_88 [7]2 years ago

8 0

Deciduous forest This is because of the location

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The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.

Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, C represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

$tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}$

The steepness of the slope is therefore;

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100$

Where;

$\overline{AM}$ = Half the width of the truck = $\dfrac{2.4 \, m}{2}$ = 1.2 m

$\overline{CM}$ = The elevation of the center of gravity above the ground = 2.2 m

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%$

$tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}$

$Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}$

The maximum steepness of the slope where the truck can be parked is 54.55 %.

Learn more here:

brainly.com/question/20793607

3 0

2 years ago

In an electrical circuit 200mC of charge flows through a 50kohm resistor in 90 seconds. What is the power

ss7ja [257]

Answer: $0.2464\ W$

Explanation:

Given

Charge $Q=200\ mC$

Resistance $R=50\ k\Omega$

Time $t=90\ s$

Charge is the amount of current flown in a particular time

$Q=It\\\Rightarrow 200\times 10^{-3}=I\times 90\\\\\Rightarrow I=\dfrac{200\times 10^{-3}}{90}\\\\\Rightarrow I=2.22\times 10^{-3}\ A$

Power is the given by $P=I^2R$

Insert values

$\Rightarrow P=\left(2.22\times 10^{-3}\right)^2\times 50\times 10^3\\\\\Rightarrow P=246.42\times 10^{-3}\ W\\\\\Rightarrow P=0.2464\ W$

The power is in the circuit is $0.2464\ W$ .

8 0

2 years ago