answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Grace [21]
2 years ago
5

A ship maneuvers to within 2500 m of an island's 1800 m high mountain peak and fires a projectile at an enemy ship 610 m on the

other side of the peak, as illustrated in Figure 3-29. If the ship shoots the projectile with an initial velocity of v = 248 m/s at an angle of θ = 74°, how close to the enemy ship does the projectile land?
Physics
2 answers:
Ne4ueva [31]2 years ago
3 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

t=(0-(250sin75)^2)/-9.8 
<span>the distance one is (2500+610)- (250m/s*cos75)*t=Dh Dh=horizontal distance </span>

<span>the max height one is d=0.5*9.8*t^2 </span>
<span>d= max height subtract 1800-d</span>
valkas [14]2 years ago
3 0

Answer: 2714.24 m

Explanation:

The projectile would have to cross the mountain peak to reach the other side and hit the enemy ship.

Vertical component of velocity = Vy = 248 sin 74° m/s = 238.4 m/s

horizontal component of velocity = Vx = 248 cos 74° m/s = 68.4 m/s

The time taken to reach the peak by the projectile would be half the time taken by projectile to cover the horizontal distance across the mountain.

Using second equation of motion:

Δy = Vy t + 0.5 a t²

⇒0 = 238.4 m/s × t + 0.5 × (-9.8 m/s²) t²

t = 238.4/4.9 = 48.6 s

Horizontal distance covered, x = Vx (t) = 68.4 m/s × 48.6 s = 3324.24 m

It lies at (3324.24 - 610) m = 2714.24 m close to the ship.

You might be interested in
The inventor of the photographic process in which a photograph produced without a negative by exposing objects to light on light
Nikolay [14]
<span>A. Man Ray --------------------</span>
7 0
1 year ago
Read 2 more answers
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
2 years ago
While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe
kaheart [24]

Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

P_{Miy} + p_{Piy} + j_{y}= P_{Mfy} +P_{pfy}

m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sinФ

60 ·8 + 0 = (60 + 80)V_{f}sinФ

480 = 140V_{f} sinФ................. (I)

y-axis component form (+y north);

P_{Mix} + p_{Pix} + j_{x} = P_{Mfx}+ P_{pfx}

m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cosФ

0 + 80.9 = (60 + 80)V_{f}cosФ

 720= 140V_{f}cosФ

140Vf=\frac{720}{cos}Ф......................................(2)

 Substituting (2) into (1) to give the angle;

 480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

480 = 140Vfsin 33.69°

Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

 

7 0
2 years ago
Determine the sign (+ or −) of the torque about the elbow caused by the biceps, τbiceps, the sign of the weight of the forearm,
Alex Ar [27]
Ans: 
1.  τbiceps = +(Positive)
2.  τforearm = -(Negative)
3.  τball = -(Negative)

Explanation:

The figure is attached down below.

1. T<span>orque about the elbow caused by the biceps, τbiceps:
Since Torque = r x F (where r and F are the vectors)
</span>Where r is the vector from elbow to the biceps.
<span>
We can see in the figure that F(biceps) is in upward direction, and by applying the right hand rule from r to F, we get the counterclockwise direction. The torque in counterclockwise direction is positive(+). Therefore, the sign would be +.

2. </span>Torque about the the weight of the forearm, τforearm:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the forearm.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(forearm) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

3. Torque about the the weight of the ball, τball:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the ball.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(ball) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

8 0
2 years ago
You are given two rectangular blocks of shiny metal, Block A and Block B, and are asked to determine which one will float in a b
vladimir2022 [97]

Answer:

Explanation:

Volume of block A = 10 x 6 x 1 = 60 cm³

Mass of block A = 630 g

density of mass A = mass / density

= 630 / 60 = 10.5g / cm³

Volume of block B = 5 x 5 x 3 = 75 cm³

Mass of block A = 604 g

density of mass A = mass / density

= 604 / 75 = 8.05 g / cm³

Since density of both A and B are less than that of mercury , both will float in mercury.

7 0
2 years ago
Other questions:
  • If steam enters a turbine at 600K and is exhausted at 400K, calculate the efficiency of the engine.
    13·2 answers
  • Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the
    11·1 answer
  • Explain how cognitive psychologists combine traditional conditioning models with cognitive processes.
    6·2 answers
  • A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the s
    11·1 answer
  • Complete each statement about the sign of the work done on a baseball. Carlton catches a baseball and his hand moves backward as
    13·1 answer
  • What is the effect of the following change on the volume of 1 mol of an ideal gas? The initial pressure is 722 torr, the final p
    13·1 answer
  • You lower the temperature of a sample of liquid carbon disulfide from 90.3 ∘ C until its volume contracts by 0.507 % of its init
    5·2 answers
  • If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
    12·1 answer
  • A sample of a gas occupies a volume of 90 mL at 298 K and a pressure of 702 mm Hg. What is the correct expression for calculatin
    9·1 answer
  • A brass lid screws tightly onto a glass jar at 20 degrees C. To help open the jar, it can be placed into a bath of hot water. Af
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!