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Arada [10]
4 years ago
9

PLEASE HELP!! PHYSICS QUIZ!

Physics
2 answers:
diamong [38]4 years ago
5 0

1.

Answer: True

Explanation: A projectile is launched at certain angle with the horizontal. Hence, its equation of motion can be described using vertical and horizontal components. Therefore, the statement is true.

2.

Answer: False

Explanation:

The shape of the trajectory of a projectile is parabola. Hence, the given statement is false.

3.

Answer: A) constant

Explanation: During the course of trajectory, the horizontal component of the velocity is constant. There is no external force acting in the horizontal direction. In the vertical direction,  acceleration due to gravity acts which increases the vertical component of velocity. Hence, the correct option is A.

Dmitriy789 [7]4 years ago
4 0

1).  Technically false if you're persnickety about your words.  But this quiz wants you to say 'True'.

The projectile doesn't have vertical and horizontal components.  But if the projectile is <em>moving</em>, then its displacement, velocity, and acceleration can have vertical and horizontal components.

2).  This one's <em>false.</em>  The shape of a projectile's trajectory is a parabola. (Unless the projectile is moving fast enough to be in a stable orbit.  THEN its path is an ellipse.)

3). During the time the projectile is flying, the horizontal component of its velocity is <em>constant</em> IF THERE'S NO AIR RESISTANCE.

If there IS air resistance, then the horizontal component of velocity is decreasing.

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7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr
Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m

d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m

From the continuity equations in pipes we have to

A_1V_1 = A_2 V_2

Where,

A_{1,2} = Cross sectional Area at each section

V_{1,2} = Flow Velocity at each section

Then replacing we have,

(\pi r_1^2) v_1 = (\pi r_2^2) v_2

(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2

v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1

From Bernoulli equation we have that the change in the pressure is

\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)

7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)

7300= 8919.01 v_1^2

v_1 = 0.9m/s

Therefore the speed of flow in the first tube is 0.9m/s

6 0
4 years ago
What is the law of motion​
777dan777 [17]

Answer:

A body at rest remains at rest and a body in motion remains in uniform motion in a straight line unless acted upon by an external force is law of motion.

5 0
3 years ago
In an attempt to reduce our dependence on non–renewable resources, and cut down on the harm to the environment, we could burn __
valentinak56 [21]

In an attempt to reduce our dependence on non-renewable resources, and cut down on the harm to the environment, we could burn biomass to produce electricity.

7 0
3 years ago
A +3.4 x 10-6 C test charge experiences forces from two other nearby charges: a 3 N force due east and a 15 N force due west. Wh
Alecsey [184]

Answer:

3.53×10⁶ N/c due west

Explanation:

From the question

E = F'/q........................ Equation 1

Where E =  Electric Field, F = Net Force, q = Charge.

But,

F' = F₂-F₁...................... Equation 2

Substitute equation 2 into equation 1

E = (F₂-F₁)/q................ Equation 3

Given: F₁ = 3 N due east, F₂ = 15 N due west,  q = 3.4×10⁻⁶ C

Substitute these values into equation 1

E = (15-3)/(3.4×10⁻⁶)

E = 12/(3.4×10⁻⁶)

E = 3.53×10⁶ N/c due west

4 0
3 years ago
Could you help me with a science question really quick?
Rama09 [41]
Yea I would love to help u
4 0
3 years ago
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