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3 years ago

Two cars approach a street corner at right angles to each

Physics

1 answer:

irakobra [83]3 years ago

4 0

Answer:

the angle is given by

Tan theta = 35/59 = 0.59

so theta = Tan ^-1 ( 0.59 )

theta = 30.54 deg.

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A 1500 kg car, initially traveling at 22.0 m/s, hits its brakes and skids to a stop. Determine the work done by friction.

Sedbober [7]

Answer:

The work done by the car is 363 kJ

Explanation:

Work : Work is said to be done when a Force moves an object through a certain distance. Work and Energy are interchangeable because they have the same unit. The unit of work is Joules (J).

Mathematically work done can be expressed as,

E = W = 1/2mv²

W = 1/2mv²................................ Equation 1

Where E = Energy, W = work done, m = mass of the car, v = velocity of the car

Given: m=1500 kg, v=22 m/s

Substituting these values into equation 1

W = 1/2(1500)(22)²

W = 750 × 484

W = 363000 J

W = 363 kJ

Thus the work done by the car is 363 kJ

8 0

3 years ago

9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the

mojhsa [17]

Answer:

Moment about SHOULDER ∑ τ = 3.17 N / m,

Moment respect to ELBOW Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

∑ τ = I α

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

∑ τ = I α

I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

I_mass = m L²

we substitute

∑ τ = (⅓ m L² + M L²) α

let's calculate

∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

Στ = I α

I = I_ forearm + I_mass

I = ⅓ m (L-0.03)² + M (L-0.03)²

let's calculate

i = ⅓ 2.3 0.47² + 0.5 0.47²

I = 0.2798 Kg m²

Στ = 0.2798 10

Στ= 2.80 N m

3 0

3 years ago

Which is true about using energy? There is an unlimited amount of energy in the universe. When energy is used, it disappears for

frutty [35]

Answer:

3. When energy is used, it can transform to new types but can never disappear.

Explanation: it can transform into heat, light, ect and will never disapear unlessed turned of.

7 0

3 years ago

Which chemical reaction shows photosynthesis

djverab [1.8K]

Answer: Option (a) is the correct answer.

Explanation:

Photosynthesis is defined as a chemical process in which plants tend to make their own food in the presence of sunlight, water and carbon dioxide.

During this process, product formed by the plants is glucose and oxygen.

The chemical reaction equation for photosynthesis is as follows.

$6CO_{2} + 6H_{2}O + \text{Sunlight} \rightarrow C_{6}H_{12}O_{6} + 6O_{2}$

Therefore, we can conclude that out of the given options the correct equation is as follows.

$\text{water + carbon dioxide + sunlight} \rightarrow \text{oxygen + glucose}$

8 0

3 years ago

Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 2.6 m long, its mass is 0.5

Irina-Kira [14]

Answer:

15.13 m/s

Explanation:

The wave speed of the stretched rope can be calculated using the following formula

$v = \sqrt{\frac{F_T}{\mu}}$

where $F_T = 44N$ is the tension on the rope and $\mu = m/L = 0.5 / 2.6 = 0.1923 kg/m$ is the density of the rope per unit length

$v = \sqrt{\frac{44}{0.1923}} = \sqrt{228.8} = 15.13 m/s$

6 0

3 years ago