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pshichka [43]
1 year ago
9

A projectile follows a straight-line path instead of a parabolic trajectory. Which could be the launch angle? would it be 90 or

45 or 3o??
Physics
2 answers:
Gnesinka [82]1 year ago
7 0
<em>projectile can only follow the straight line path when it is launched upward straightly so the correct option is <u>90 degree with respect to horizontal x -axis ..:)</u></em>
Alja [10]1 year ago
6 0

Answer:

90 degree

Explanation:

For the horizontal projection or the vertical projection, the path traversed by the projectile is a straight line .

So the angle of projection is 0 degree or 90 degree.

In this condition the motion is in one dimension. If the angle of projection is any arbitrary angle except 0 degree or 90 degree, the motion of the projectile is parabolic.

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In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
jekas [21]

Answer:

The weight of heart of a human is 0.93 lbs.

Explanation:

Given that,

Approximately weight of heart is 0.5 % of the total body weight.

Weight of human = 185 lbs

Let the the weight of total body is w and weight of heart is w_{h}.

We need to calculate the weight of heart of a human

Using given data

w_{h}=0.5\times w

Where, h = weight of heart

w = weight of human

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

Hence, The weight of heart of a human is 0.93 lbs.

8 0
2 years ago
A shot-putter exerts a force of 0.142 kN on a shot, accelerating it to 22.75 m/s2. What is the mass of the shot?
Svetach [21]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Force and Power.

Since, according to the Newton's law,

Force = mass * Acceleration.

hence, here

Force = 142 N, accelration = 22.75 m/s2

hence, mass = 142/22.75

===> Mass = 6.24 Kg

hence the mass of the shot is 6.24 Kg

8 0
1 year ago
A piston-cylinder chamber contains 0.1 m3 of 10 kg R-134a in a saturated liquid-vapor mixture state at 10 °C. It is heated at co
vaieri [72.5K]

Answer:

(A) 10132.5Pa

(B)531kJ of energy

Explanation:

This is an isothermal process. Assuming ideal gas behaviour then the relation P1V1 = P2V2 holds.

Given

m = 10kg = 10000g, V1 = 0.1m³, V2 = 1.0m³

P1 = 101325Pa. M = 102.03g/mol

P2 = P1 × V1 /V2 = 101325 × 0.1 / 1 = 10132.5Pa

(B) Energy is transfered by the r134a in the form of thw work done in in expansion

W = nRTIn(V2/V1)

n = m / M = 10000/102.03 = 98.01mols

W = 98.01 × 8.314 × 283 ×ln(1.0/0.1)

= 531kJ.

6 0
2 years ago
Which of the following sketches represents a possible configuration for this problem?
garri49 [273]
Where are the following sketches?
7 0
2 years ago
The 1.5-in.-diameter shaft AB is made of a grade of steel with a 42-ksi tensile yield stress. Using the maximum-shearing-stress
PolarNik [594]

Answer:

T = 0.03 Nm.

Explanation:

d = 1.5 in = 0.04 m

r = d/2 = 0.02 m

P = 56 kips = 56 x 6.89 = 386.11 MPa

σ = 42-ksi = 42 x 6.89 = 289.58 MPa

Torque = T =?

<u>Solution:</u>

σ = (P x r) / T

T = (P x r) / σ

T = (386.11 x 0.02) / 289.58

T = 0.03 Nm.

7 0
2 years ago
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