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sashaice [31]
3 years ago
8

A projectile fired up into the air at an angle has a range of 235 m and a flight time of 47 s.

Physics
1 answer:
madam [21]3 years ago
7 0
<h2>Answer:5ms^{-1},133.6m,51.18ms^{-1}</h2>

Explanation:

Let v_{x},v_{y} be the horizontal and vertical components of velocity.

Question a:

Horizontal component of velocity is the ratio of range and time of flight.

So,horizontal component of velocity is \frac{235}{47}=5ms^{-1}

So,v_{x}=5ms^{-1}

Question b:

Time of flight=\frac{2v_{y}}{g}

So,v_{y}=\frac{47\times 9.8}{2}=51.18ms^{-1}

Maximum height is given by \frac{v_{y}^{2}}{2g}

So,maximum height is \frac{51.18^{2}}{2\times 9.8}=133.6m

Question c:

The vertical velocity is already calculated in Question b.

v_{y}=51.18ms^{-1}

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Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge 15\times10^-6 C is placed which is the origin.

Let B be the point where the charge 9\times 10^-6 C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, \frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2},

Q_A = 15\times 10^-6 C

Q_B=9\times10^-6C

d_A = 1+x cm

d_B=x cm

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

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1 year ago
How thick is the Earth's surface?
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It is typically 30 km to 50 km thick.
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As the distance between two objects increases, the gravitational force of attraction between them will
jeka94
That force DEcreases. This is a big part of the reason why the Earth attracts you with more force than Jupiter does.
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3 years ago
A particular 12 V car battery can send a total charge of 110 A·h (ampere-hours) through a circuit, from one terminal to the othe
DiKsa [7]
<h2>Answer:</h2>

(a) 3.96 x 10⁵C

(b) 4.752 x 10⁶ J

<h2>Explanation:</h2>

(a) The given charge (Q) is 110 A·h (ampere hour)

Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;

=> Q = 110A·h

=> Q = 110 x 1A x 1h          [1 hour = 3600 seconds]

=> Q = 110 x A x 3600s

=> Q = 396000A·s

=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C

Therefore, the number of coulombs of charge is 3.96 x 10⁵C

(b) The energy (E) involved in the process is given by;

E = Q x V           -----------------(i)

Where;

Q = magnitude of the charge = 3.96 x 10⁵C

V = electric potential = 12V

Substitute these values into equation (i) as follows;

E = 3.96 x 10⁵ x 12

E = 47.52 x 10⁵ J

E = 4.752 x 10⁶ J

Therefore, the amount of energy involved is 4.752 x 10⁶ J

8 0
3 years ago
Given a force of 100 N and a acceleration of 5 m/s, what is the mass
SIZIF [17.4K]

Answer:

force = mass \times acceleration \\ 100 = m \times 5 \\ m =  \frac{100}{5}  \\ m = 20 \:  kg

8 0
2 years ago
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