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gayaneshka [121]
4 years ago
5

Which image depicts projectile motion?

Physics
2 answers:
Elis [28]4 years ago
6 0

Answer:

the answer is the second one

Explanation:

the person is moving through the air while the other people are either not moving forward or backwards or on the ground with full controll over where they go

tia_tia [17]4 years ago
4 0

Answer:

Second picture is representing projectile motion

Explanation:

As we know that in projectile motion one of the component of the motion of object will remains constant while other component which is vertical in direction will change due to gravity.

So here in this type of motion the path of the object is same as the parabolic path.

Here we know that vertical component of the motion will change due to gravity while horizontal component will remain constant always

So here we can say that

correct image is second image

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Plz help me ! I hate apex
statuscvo [17]

Answer:

1. Observation

2. Data collection

3. Hypothesis

4. Experiment

5. Result

6. Conclusion

7. Theory

8. Law

Explanation:

4 0
4 years ago
Read 2 more answers
A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show
Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]

And the speed can be calculated as follows:

v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]

7 0
3 years ago
Help with 1 2 and 3 please
geniusboy [140]

1. Amperes, is the SI unit (also a fundamental unit) responsible for current.

2. I = \frac{q}{t} Δq over Δt technically

Rearrange for Δq

I x Δt = Δq

1.5mA x 5 = Δq

Δq = 0.0075

Divide this by the fundamental charge "e"

Electrons: 0.0075 / 1.60 x 10^-19

Electrons: 4.6875 x 10^16 or 4.7 x 10^16

3.  So we know that the end resistances will be equal so:

ρ = RA/L

ρL = RA

ρL/A = R

Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

\frac{p1L1}{A1}  = \frac{p2L2}{A2}\\

We are looking for L2 so we can isolate using algebra to get:

\frac{A2(\frac{P1L1}{A1}) }{P2} = L2

If you fill in those values you get 0.0205

or 2.05 cm



6 0
3 years ago
Some persons are counted as out of the labor force because they have made no serious or recent effort to look for work. However,
Sindrei [870]
Both the employment rate and labor force participation rate would be higher
3 0
3 years ago
They did a flu shot from McKinnon at 45° to the horizontal with an initial speed of 25 m/s and that is positioned at a horizonta
Margarita [4]

Answer:

The time taken for the daredevil to travel the 50 m horizontally is 2.83 s.

Explanation:

Given;

angle of projection, θ = 45°

initial speed of the projectile, u = 25 m/s

horizontal distance traveled by the projectile, x = 50 m

The time taken for the daredevil to travel the 50 m horizontally is calculated as;

t =\frac{X}{u_x}

where;

u_x is the horizontal component of the velocity = uCosθ

t =\frac{X}{u Cos \ \theta} \\\\t =\frac{50}{25 \times Cos(45)} \\\\t= \frac{50}{17.678} \\\\t = 2.83 \ s

Therefore, the time taken for the daredevil to travel the 50 m horizontally is 2.83 s.

6 0
3 years ago
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