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deff fn [24]
3 years ago
5

Ocean waves that measure 12m from crest to adjacent crest pass by a fixed point every 2.0s what is the speed of the waves

Physics
1 answer:
Darina [25.2K]3 years ago
7 0
Because there is a 1.2m distance between the crest and the adjacent trough in the series. 
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Please help me out as soon as you can. Really need help.
olga nikolaevna [1]
The answer would be 6 because 2.0x3= 6

(newton’s 2nd law)

mark me brainliest
4 0
3 years ago
A object travels at constant negative acceleration. What does the graph of the object's velocity as a fun
melomori [17]
Constant = straight line
“Travels at constant negative acc.”
Which is negative slope

Solution: B. Straight line w/ neg. slope
7 0
3 years ago
The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0
3 years ago
An object with a mass of 6.0 kg accelerates 8.0 m/s^2 when an unknown force is applied to it. What is the amount of the force? R
iren [92.7K]

Answer:

48N

Explanation:

use F=ma, or force is equal to mass multiplied by acceleration.

8 0
3 years ago
What kind of friction occurs as a fish swims through water?
Paul [167]
It must be sliding friction, because the fish is already in motion.
7 0
3 years ago
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