Can i get the answer quickly plz

Aleks04 [339]

Answer:

Please find detailed explanation of second class levers below

Explanation:

Levers are one of the classes of machine that possesses three levels namely: first class, second class and third claas. A second class lever is the level of levers in which the load (L) is in between the pivot (F) and the effort (E).

Examples of second class levers include; wheelbarrow, a bottle opener etc. In the bottle opener for example, the bottle lid (load) is in between the pivot of the opener and the hand opening it (effort).

6 0

3 years ago

Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o

seraphim [82]

Answer:

$\theta = 15^o \: or\: 75^o$

Explanation:

As we know that the formula of range is given as

$R = \frac{v^2sin2\theta}{g}$

now we know that

maximum value of the range of the projectile is given as

$R_{max} = \frac{v^2}{g}$

now we need to find such angles for which the range is half the maximum value

so we will have

$\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}$

$sin(2\theta) = \frac{1}{2}$

$2\theta = 30 or 150$

$\theta = 15^o \: or\: 75^o$

7 0

3 years ago

You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m

sergey [27]

Answer:

$v=12.65\ m.s^{-1}$

Explanation:

Given:

mass of vehicle, $m=1350\ kg$

radius of curvature, $r=71\ m$

coefficient of friction, $\mu=0.23$

<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

$m.\frac{v^2}{r} =\mu.N$

where:

$\mu=$ coefficient of friction

$N=$ normal reaction force due to weight of the car

$v=$ velocity of the car

$1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)$

$v=12.65\ m.s^{-1}$ is the maximum velocity at which the vehicle can turn without skidding.

5 0

3 years ago

What happens as water evaporates from a lake

NARA [144]

Answer:

As that liquid water is further heated, it evaporates and becomes a gas—water vapor. So C... Need branliest pls

Explanation:

4 0

3 years ago

A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under

Kaylis [27]

Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ = (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] = 1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ = 1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)] = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ = (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

4 0

3 years ago