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2 years ago

2. A body is thrown vertically upward with a speed of 100 m/s.The time taken to be

Physics

1 answer:

Pachacha [2.7K]2 years ago

4 0

Answer:

b. 20 sec

Explanation:

y = y₀ + v₀ t + ½ g t²

0 = 0 + (100) t + ½ (-10) t²

0 = 100t − 5t²

0 = t (100 − 5t)

t = 0, t = 20

The body lands after 20 seconds.

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A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0

2 years ago

The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v

velikii [3]

Answer:

$\theta = n\pi/2, {\rm where~n~is~an~integer.}$

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

$\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k$

Similarly,

$\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k$

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

$\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)$

So,

$\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)$

At time t = 0, this equation becomes

$0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}$

7 0

3 years ago

Because of your success in physics class you are selected for an internship at a prestigious bicycle company in its research and

tiny-mole [99]

To develop the problem it is necessary to apply the equations related to the moment of inertia.

The given values can be defined as,

$M = 1.0 kg$

$r = 0.5 m$

$m = 10 g$

$I = 0.280 kg.m^2$

According to the definition of the moment of inertia applied to the exercise we can arrive at the equation that,

$I = I_{rim} + n * I_{spoke}$

Where n is the number of spokes necessary to construct the wheel.

$I_{rim} = M*r^2 = 1.0 * 0.5^2$

$I_{spoke} = \frac{1}{3} * m * r^2 = \frac{1}{3}* 10 * 10^-3 * 0.5^2$

Replacing the values at the general equation we have,

$0.280 = 1.0 * 0.5^2 + n * (1/3 * 10 * 10^-3 * 0.5^2 )$

Solving for n,

$n = 36$

Therefore the number of spokes necessary to construct the wheel is 36

PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then

$M_w= M + n*m$

$M_w = 1.0 + 36* 10 * 10^{-3} Kg$

$M_w = 1.36 Kg$

Therefore the mass of the wheel must be of 1.36Kg

4 0

3 years ago

Hanna tosses a ball straight up with enough speed to rermain in the air for several seconds

Ipatiy [6.2K]

What a relief ! That gives her time to step out of the way, before the ball
comes crashing down in the same place where she was standing.

7 0

3 years ago

What effect did printing with movable type have on people during the Renaissance?

vodka [1.7K]

Answer: He didn't invent printing. He didn't even invent movable type. He often ran into legal trouble and, when he died in 1468, he did so with little money or glory.

Explanation:

3 0

3 years ago