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Montano1993 [528]

3 years ago

14

The y-component of a projectile’s velocity is 12.1 m/s. When the projectile once again passes by the height from which it was la

unched, what is the y-component of its velocity? impossible to calculate without knowing the x-component 0 m/s 12.1 m/s -12.1 m/s

1 answer:

Nat2105 [25]3 years ago

6 0

It's 12.1 m/s, assuming that's the launch velocity that's given.
For projectile motion, velocity's y-component is parabolic/quadratic. It's x-component is constant, so you don't need to know it.

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What kind of circuit is the one shown below?

lions [1.4K]

Answer:

The given circuit diagram shows parallel circuit.

Explanation:

In this circuit diagram two bulbs are connected in parallel combination because current flows from the battery gets bifurcated at the junction. Thus, two bulbs are connected in parallel combination.

This parallel combinations of bulbs then connected to the battery given in the diagram. So, the combinations of bulbs are connected in parallel combinations with the battery.

Hence, both bulbs and battery are connected in parallel combinations with each other.

The circuit diagram shown in figure is parallel.

8 0

3 years ago

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Question 4 How much time does it take to walk 8 km north at a velocity of 3.8 km/h?

IrinaVladis [17]

Given parameters:

Displacement = 8km

Velocity = 3.8km/h

Unknown:

time = ?

Solution:

Velocity is displacement divided by time.

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Displacement = velocity x time

Input the parameters:

8 = 3.8 x time

Time = $\frac{8}{3.8}$ = 2.1s

The time taken is 2.1s

6 0

3 years ago

Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v

d1i1m1o1n [39]

1) A) Object 1 has the greater momentum

The magnitude of the momentum of an object is given by:

$p=mv$

where

m is the mass of the object

v is its speed

Object 1 has a mass of $m_1 = 2m$ and a speed of $v_1 = v$ , so its momentum is

$p_1 = m_1 v_1 = (2m)(v)=2mv$

Object 2 has a mass of $m_2 = m$ and a speed of $v_2 = \sqrt{2} v$ , so its momentum is

$p_2 = m_2 v_2 = (m)(\sqrt{2} v)=\sqrt{2}mv$

So we see that $p_1 > p_2$ , so object 1 has the greater momentum.

2) The objects have the same kinetic energy.

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

Object 1 has a mass of $m_1 = 2m$ and a speed of $v_1 = v$ , so its kinetic energy is

$K_1 = \frac{1}{2}m_1 v_1^2 = \frac{1}{2}(2m)(v)^2=mv^2$

Object 2 has a mass of $m_2 = m$ and a speed of $v_2 = \sqrt{2} v$ , so its kinetic energy is

$K_2 = \frac{1}{2}m_2 v_2^2 = \frac{1}{2}(m)(\sqrt{2} v)^2=mv^2$

So we see that $K_1 =K_2$ , so the objects have same kinetic energy

5 0

3 years ago

For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103

Alona [7]

Answer:

a) P = 44850 N

b) $\delta l =0.254\ mm$

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

$\sigma=\frac{\textup{Load}}{\textup{Area}}$

on substituting the values, we get

$345\times10^6=\frac{\textup{Load}}{0.00013}$

or

Load, P = 44850 N

Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>

b)The deformation ( $\delta l$ ) due to an axial load is given as:

$\delta l =\frac{PL}{AE}$

on substituting the values, we get

$\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}$

or

$\delta l =0.254\ mm$

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3 years ago

What types of bonds do you think the good conductors of electricity have?

mariarad [96]

This what i found hope it helps

6 0

3 years ago