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Dmitry_Shevchenko [17]
2 years ago
12

Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at t

he point of ejection are 1 meter/second and 0.5 meters/second respectively, calculate the ratio of their horizontal distances of impact.
Physics
2 answers:
IrinaK [193]2 years ago
4 0
 t1 = √2h1/g = √2*0.5/9,8 = 0.319 sec 
t2 = √2h2/g = √2*1.0/9,8 = 0.451 sec 

In which t = times for the vertical movement
h = height
g= gravity (we use standardized measurement of 9.8)

d1 = 1*0.319 = 0.319 m 
d2 = 0.5*0.451 = 0.225 m 

in which d = Horizontal distance

ratio
= di : d2
= 0.319 : 0.225    

 = 3.19 : 2.25
Lubov Fominskaja [6]2 years ago
3 0

Answer:

1.41:1

Explanation:

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2 years ago
Read 2 more answers
A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when
choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

F_t - mg = ma

here since speed is constant or it is at rest

so we will have

a = 0

F_t = mg

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

F_t - mg = ma

F_t = mg + ma

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

F_t - mg = -ma

F_t = mg - ma

so tension force is less than the weight of crate if it is accelerating downwards

4 0
3 years ago
A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
3 years ago
An object is placed at zero zero on a Number line. It moves three units to the right, then four units to the left, and then 60 u
STatiana [176]

Answer:

the displacement of the object is 5 units

Explanation:

The computation of the displacement of the object is shown below:

= Move to the right + move to the right - move to the left

= 6 units + 3 units - 4 units

= 9 units - 4 units

= 5 units

Hence, the displacement of the object is 5 units

7 0
2 years ago
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2 years ago
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