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2 years ago

A projectile is fired at an angle such that the vertical component of its velocity

Physics

1 answer:

mestny [16]2 years ago

3 0

Answer:

It takes 3 seconds to reach its high point

Explanation:

Projectile Motion

In a projectile motion (or 2D motion), the object is launched with an initial angle θ and an initial velocity vo.

The components of the velocity are

$v_{ox}=v_o\cos\theta$

$v_{oy}=v_o\sin\theta$

The speed in the horizontal direction at any time t is:

$v_y=v_o\sin\theta-g.t$

The time taken to reach the maximum height is when vy=0, or:

$\displaystyle t_m=\frac{v_o\sin\theta}{g}$

We are given the y-component of the velocity, thus:

$\displaystyle t_m=\frac{30~m/s}{10~m/s^2}$

$t_m=3~s$

It takes 3 seconds to reach its high point

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Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 15.0 ∘ west of north,

Darina [25.2K]

Answer:

The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

Explanation:

The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.

The angle between the 2 forces and displacement is ∅ = 15°.

First we have to calculate the work done by the individual force and then we can calculate the total work.

The work done on a particle by a constant force F during a straight line displacement s is given by following formula:

W = F*s

W = F*s*cos∅

With ∅ = the angles between F and s

The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N

The total work done can be calculated as followed:

Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2

Wtotal = 2Fs*cos∅

Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°

Wtotal = 3.44*10^9 Nm = 3.44 *10^9 J

The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

5 0

2 years ago

The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n

monitta

Answer:

i know the questin but i got to try and find it

Explanation:

5 0

2 years ago

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

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6 0

1 year ago

the Space Program has benefitted people in everyday life. Describe two ways in which people can benefit

ycow [4]

So we can know what is in space maybe weird or interesting stuff

8 0

3 years ago

A student drops two metallic objects into a 120-g steel con- tainer holding 150 g of water at 25°C. One object is a 200-g cube o

marissa [1.9K]

Answer:

The mass of the aluminum chunk is 258 g

Explanation:

Given;

mass of steel container = 120-g

mass of water = 150 g

initial temperature of water, = 25°C

mass of copper cube, $M_{cu}$ = 200 g

initial temperature of the copper cube, $T_c_u$ = 85°C

initial temperature of the aluminum chunk $T_A_l$ = 5.0°C

Neglecting heat loss, heat exchanged by the two metallic objects is the same since initial temperature is equal to final temperature of water.

$M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu}$

where;

$C_{AL}$ is specific heat capacity of aluminum

$\delta T_{Al}$ is change in temperature of aluminum

$C_c_u$ is the specific heat capacity of copper

$\delta T_c_u$ is the change in temperature of copper

$M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu} \\\\M_{Al} = \frac{M_{cu}C_{cu} \delta T_{cu}}{C_{Al} \delta T_{Al}} \\\\M_{Al} = \frac{0.2*387*60}{900*20} = 0.258 \ kg$

Therefore, the mass of the aluminum chunk is 258 g

7 0

3 years ago