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marin [14]
2 years ago
9

1- Show that the equation d=vt2 dimensionally correct or not?

Physics
1 answer:
TiliK225 [7]2 years ago
5 0
- Show that the equation d=vt2 dimensionally correct or not? ... nalsn7 is waiting for your help. Add your answer and earn points.
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Which trophic level has the least available energy in kilojoules in this food web?
insens350 [35]

The highest trophic level has the least available energy in kilojoules.

Even though the food web is not shown in the question, but we know that energy decreases steadily as it is passed on from one trophic level to the next according to the second law of thermodynamics.

Energy enters into the system from the sun. The primary producers utilize this energy to produce food. As plants are eaten by animals, this energy is transferred along the food web an diminishes at each higher trophic level.

At the highest trophic level, the the least available energy in kilojoules in this food web is found.

Learn more: brainly.com/question/2233704

7 0
2 years ago
.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c
ohaa [14]

Answer:

 a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

          I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

        F t = 2 (m v)

therefore the average force is

         F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

        Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

        Em_f = K = ½ m v²

energy is conserved in this system

        Em₀ = Em_f

        m g h = ½ m v²

        v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

        F = 2m √(2gh)   /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

          F = m a

          a = F / m

 

          a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

         a = 2 √ (2 9.8 1.5) / 10⁻³

         a = 1.1 10⁵ m / s²

6 0
3 years ago
Calculate the potential energy of a 5 kg object sitting at the top of a 2 meter ramp.
trapecia [35]
The formula used to find potential energy is <em>P.E. = M * G * H</em> (P.E. is potential energy, M is mass, G is gravitational pull, and H is height). So the answer to your question is <em>5 * 9.8 * 2</em>, which equals 98.
5 0
3 years ago
A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the
alex41 [277]

Answer:

(a)v_1 = a_1t_1 = 1.76 t_1

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1

(b) The velocity of the car before the driver begins braking is

v_1 = 1.76*20 = 35.2m/s

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2

We can use the following equation of motion to calculate how far the car has travel since braking to stop

s_2 = v_1t_2 + a_2t_2^2/2

s_2 = 35.2*5 - 7.04*5^2/2 = 88 m

Also the distance from start to where the driver starts braking is

s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0
3 years ago
​Hooke's Law. The distance d when a spring is stretched by a hanging object varies directly as the weight w of the object. If th
nadezda [96]

Answer:

distance when the weight is 8 ​kg is 26.66 cm

Explanation:

given data

distance d2 = 10 cm

weight w2 = 3 ​kg

weight w1 = 8 kg

to find out

distance when the weight is 8 ​kg

solution

we consider here distance d1 when weight is 8 kg

so equation will be

d1/d2 = w1/w2

d/10 = 8/ 3

so d = 8/3 × 10

so d = 26.66

distance when the weight is 8 ​kg is 26.66 cm

7 0
3 years ago
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