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2 years ago

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Physics

2 answers:

GaryK [48]2 years ago

5 0

Distance traveled=(80 km/1hr)*(2.5 hrs/1) or 200 km.

Does This Help?

ycow [4]2 years ago

5 0

Answer:

200 kilometers

Explanation:

Distance: Speed × Time taken

Here given:

a. speed: 80 km/h or 80 km per hour

b. time taken: 2.5 hours

Solve by applying formula:

distance = speed × time taken = 80 × 2.5 = 200 km

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A point P is placed exactly between two charges, Q1 and Q2. If the electric field experienced by point P due to charge Q1 is 1.5

iren2701 [21]

Answer: D

Given data:

The electric field due to charge q₁ is (E₁) = 1.5 × 10⁵ N/C

The electric field due to charge q₂ is (E₂) = 7.2 × 10⁵ N/C

Determine the net electric field at point P (Enet) = ?

electric field measured in Newtons/coulomb

We know that,

The net electric field (Enet) is vector sum of E₁ and E₂ caused by the charge q₁ and q₂ respectively

Enet = E₁ + E₂

= (1.5 × 10⁵)+ (7.2 × 10⁵)

= 8.7 × 10⁵ Newtons/coulomb

The net electric field at point P is 8.7 × 10⁵ N/C

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4 years ago

Monochromatic light with wavelength 590 nm passes through a single slit 2. 30 ?m wide and 1. 90 m from a screen. Find the distan

aivan3 [116]

Answer:

The fringes are 4.7*10^-7 m apart, such that they are adjacent.

Explanation:

Using the formula for adjacent fringes given a single slit:

Δ $x=\frac{(Wavelength)(Distance between slit and screen)}{Width}$

Δ $x=\frac{(590/10^{9})(1.90) }{(2.30)}$

Δ $x=0.000000487 m$

Hope this helps!

7 0

2 years ago

Read 2 more answers

An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.

iren2701 [21]

The motion of the package can be described as the motion of a projectile,

given that it has an horizontal velocity and it is acted on by gravity.

a) $\vec{v}$ = 70·i

b) The package will reach ground in approximately 12.77 seconds.

c) The speed of the package as it lands is approximately 145.51 m/s.

d) The path of the package based on a stationary frame of reference is parabolic

e) The path of the package as seen from the plane is directly vertical downwards

Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, $\vec{v}$ = 70·i

b) The time it will take the package to reach the ground, t, is given by the formula;

$\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}$

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

$\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}$

Which gives;

$\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}$

The time it will take the package to reach the ground, t ≈ 12.77 seconds

c) The vertical velocity just before the package reaches the ground, $v_y$ , is given as follows;

$v_y^2$ = 2·g·h

Therefore;

$v_y$ = √(2·g·h)

Which gives;

$v_y$ = √(2 × 9.81 × 800) ≈ 125.28

$v_y$ ≈ 125.28 m/s

Which gives; $\vec{v}$ = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

The speed of the package as it lands, |v| ≈ 143.51 m/s

d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a parabolic shape.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating directly vertical

downwards.

Learn more about projectile motion here:

brainly.com/question/1130127

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Vladimir79 [104]

Kinetic energy is equal to half of an objects mass multiplied by the velocity squared.

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iragen [17]

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4 years ago