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Morgarella [4.7K]
3 years ago
14

The length of a retanle is twice the width plus 10. if the perimenter of the rectangle must be 200 inches what is the maimumarea

of the rectange?
Physics
1 answer:
timurjin [86]3 years ago
5 0
<h2>Area = 2100 inch²</h2>

Explanation:

The length of a rectangle is twice the width plus 10

Let L be the length and W be the width.

               L = 2W + 10

The perimeter of the rectangle must be 200 inches

           Perimeter = 2 (L+W) = 200

                              2(2W+10+W) = 200

                                 3W + 10 = 100

                                 3W = 90

                                    W = 30 inches

                 L = 2 x 30 + 10 = 70 inches

Area = Length x Width

Area = 30 x 70

Area = 2100 inch²

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Zina [86]

Answer:

See below...

Explanation:

Let’s express ⟨α⟩ in terms of ωi , ωf , and Δt. and torque in terms of It , ωi , ωf , and Δt.

STEP 1.  

The rate of change of angular velocity is Angular acceleration.  

The net change in angular velocity is Average angular acceleration divided by the elapsed time.

⟨α⟩ = ω f −ω i/Δt

STEP 2.

Torque is assumed this way

          dω

   τ = I ----

           dt

.

⟨τ ⟩ = I t (ω f −ω i )/Δt

6 0
3 years ago
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
3 years ago
A car of mass 2200 kg collides with a truck of mass 4500 kg, and just after the collision the car and truck slide along, stuck t
katovenus [111]

Answer:

37.7m/s: principle of conservation of momentum

Explanation:

The principle to make use of is the principle of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of momentum of bodies after collision. This bodies will move with the same velocity after collision.

Momentum = Mass × velocity

For car of mass 2200kg moving with velocity 33m/s:

Momentum of car before collision = 2200×33

= 72,600kgm/s

For the truck of mass 4500kg;

Momentum = 4500 ×(22-(-18)

= 4500×40

= 180000kgm/s

After collision, their momentum is:

Momentum after collision = (2200+4500)v

= 6700v

Using the principle above to get the common velocity v we have

72600+180000 = 6700v

252600 = 6700v

v = 252600/6700

v = 37.7m/s

7 0
3 years ago
An automobile traveling at 25.0 km/h along a straight, level road accelerates to 65.0 km/h in 6.00 s. what is the magnitude of t
USPshnik [31]
Note that
1 km/h = (1000 m)/(3600 s) = 0.27778 m/s

Initial velocity, v₁ = 25 km/h = 6.9444 m/s
Final velocity, v₂ = 65 km/h = 18.0556 m/s

Time interval, dt = 6 s.

Calculate average acceleration.
a = (v₂ - v₁)/dt
   = (18.0556 - 6.9444 m/s)/(6 s)
   = 1.852 m/s²

Answer:
The average acceleration is 1.85 m/s² (nearest hundredth)
3 0
3 years ago
An Arrow (0.5 kg) travels with velocity 20 m/s to the right when it pierces an apple (2 kg) which is initially at rest. After th
sattari [20]

Answer:

4 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')...................... Equation 1

Where m = mass of the arrow, u = initial velocity of the arrow, m' = mass of the apple, u' = initial velocity of the apple, V = Final velocity of the apple and the arrow after collision.

make V the subject of the equation

V = (mu+m'u')/(m+m').................... Equation 2

Given: m = 0.5 kg, m' = 2 kg, u = 20 m/s, u' = 0 m/s(initially at rest)

Substitute into equation 2

V = (0.5×20+2×0)/(2+0.5)

V = 10/2.5

V = 4 m/s.

Hence the final velocity of the apple and the arrow after the collision = 4 m/s

6 0
3 years ago
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