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kompoz [17]
2 years ago
12

True or False?waves transfer the material from one location to another​

Physics
1 answer:
MArishka [77]2 years ago
3 0

Answer: False

Explanation:

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A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i
vovangra [49]

Answer:

c=71.4m/s

\theta=8.54\textdegree

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed V_w= 15 m/s \angle  45 \textdegree

Generally Resolving vector mathematically

  sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6

Generally the equation Pythagoras theorem is given mathematically by

c^2=a^2+b^2

c^2=10.6^2 +(10.6+60)^2

c=\sqrt{10.6^2 +(10.6+60)^2}

Therefore Resultant velocity (m/s)

c=71.4m/s

b)Resultant direction

Generally the equation for solving Resultant direction

\theta=tan^-1(\frac{y}{x})

Therefore

\theta=tan^-1(\frac{10.6}{70.6})

\theta=8.54\textdegree

7 0
2 years ago
Câu 1. Trường hợp nào dưới đây không phải là vật sáng?
Marianna [84]

Answer:

A

Explanation:

A. The pencil is on the table in broad daylight

5 0
2 years ago
In a convex lense f=20cm, m=1,then what is u and v?​
katen-ka-za [31]

Answer:

I'm not sure if I know whatever the answer is

3 0
2 years ago
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di
Pachacha [2.7K]

Answer:

m=146.277kg which is rounded to 146kg

Explanation:

Remember that F=ma

But F represents not 250N, but 250cos(35)N since the force is being pulled above the horizontal.

So 250cos(35)=204.7880111 approximately, and since a=1.4m/s^2, we have 204.7880111=m(1.4m/s^2). Then we divide both sides by the acceleration to get the mass. So m=146.2771508kg which the nearest number is 146kg

Mass is always in kg, unless stated otherwise.

4 0
2 years ago
Read 2 more answers
Can someone help me?​
Leviafan [203]

Car X traveled 3d distance in t time.  Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t,  the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

V_y.t = \frac{2d}{t} .t = 2d

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

4 0
2 years ago
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