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OlgaM077 [116]
3 years ago
11

A 216-m-wide river flows due east at a uniform speed of 2.2 m/s. A boat with speed of 7.9 m/s relative to the water leaves the s

outh bank of the river pointed in a direction 34 degrees west of north. What is the magnitude of the boat's velocity relative to the ground (in m/s)
Physics
1 answer:
Mademuasel [1]3 years ago
7 0

Answer:

The magnitude of the velocity of boat w.r.t ground is 6.2015 m/s

Solution:

As per the question;

Width of the river, w = 216 m

Uniform speed of boat due east, u_{x} = 2.2\hat{i} m/s

Angle in north west direction, \theta = 34^{\circ}

Velocity of the boat relative to water, v_{bw} = 7.9m/s

Now,

Velocity of the boat relative to the water is given by:

v_{bw} = -7.9cos34^{\circ}\hat{i} + 7.9sin34^{\circ}\hat{j}

v_{bw} = - 6.55\hat{i} + 4.42\hat{j} m/s

Also, velocity of water w.r.t ground is v_{wg} = u_{x} = 2.2\hat{i} m/s

Now,

Magnitude of the velocity of boat w.r.t ground is given by:

v_{bg} = v_{bw} + v_{wg}

v_{bg} = - 6.55\hat{i} + 4.42\hat{j} + 2.2\hat{i}

v_{bg} = - 4.35\hat{i} + 4.42\hat{j} m/s

|v_{bg}| = \sqrt{(- 4.35)^{2} + (4.42)^{2}} = 6.2015\ m/s

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sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N
yaroslaw [1]

Answer:

a)-2m/s^2

b)27.2m/s

Explanation:

Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)

W=455N=weight

W=mg

W=455N=weight

m=\frac{W}{g} =\frac{455}{9.81}=46.38kg

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

a=\frac{T-W}{m} =\frac{361-455}{46.38kg} =-2m/s^2

for point b we use the equations of motion with constant acceleration to find the velocity

Vf=\sqrt{X(2)(a)+Vo^2}

Where

Vf = final speed

Vo = Initial speed =0

A = acceleration =2m/s

X = displacement =6.8m

Solving

Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s

4 0
3 years ago
a wave is described by where x is in meters, y is in centimeters and t is in seconds. The angular frequency is
Sergeeva-Olga [200]

Complete Question

A wave is described by y(x,t) = 0.1 sin(3x + 10t), where x is in meters, y is in centimetres and t is in seconds. The angular wave  frequency is

Answer:

The  value is w =  10 \ rad /s

Explanation:

From the question we are told that  

    The equation describing the wave is y(x,t) = 0.1 sin(3x + 10t)

Generally the sinusoidal equation representing the motion of a wave is mathematically represented as

         y(x,t) =  Asin(kx + wt )

Where  w  is the  angular frequency

Now comparing this equation  with that given we see that

       w =  10 \ rad /s

 

               

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balandron [24]
It reaches 10 or 20 million degrees kelvin but it can get as high as 10 million degrees kelvin
5 0
3 years ago
Just need help with 1 and 2 please :D i’m having a bit of trouble :/
dexar [7]
1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.

2. A) 400m
B)0m
C)d=1/2(vi+vf)t
400=1/2(0+vf)92
8.7m/s
D) 0m/s
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Plz like if it helped.
7 0
3 years ago
Transform boundaries are classified under which type of fault?
cluponka [151]

Answer:

Strike-slip fault

Explanation:

Transform boundaries play the role of connecting the other plate boundary segments.

When the plates are rubbed against each other, they result in enormous amount of stresses which leads to the breaking of the part of a rock causing earthquakes. Places of occurrence of these breaks are termed as faults.

Strike slip faults results from compression which takes place horizontally, but but in this the rock displacement  releases energy and takes place in a horizontal direction which is parallel to the force of compression.

8 0
3 years ago
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