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s344n2d4d5 [400]
3 years ago
13

A projectile has an initial horizontal velocity of 34.0 M/s at the edge of a roof top. Find the horizontal and vertical componen

ts of its velocity after 5.5s
I need shown work !
Physics
1 answer:
Sveta_85 [38]3 years ago
8 0

Answer:

v_x=34 m/s

v_y=53.9\ m/s

Explanation:

<u>Horizontal Launch</u>

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

vx=v

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

v_y=g.t

The horizontal component of the velocity is always the same:

v_x=34 m/s

The vertical component at t=5.5 s is:

v_y=9.8*5.5=53.9

v_y=53.9\ m/s

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amm1812

A) position time graph for both is shown

here one of the graph is of lesser slope which means it is moving with less speed while other have larger slope which shows larger speed

At one point they intersects which is the point where they both will meet

B) Let the two will meet after time "t"

now we can say that

if they both will meet after time "t"

then the total distance moved by you and other person will be same as the distance between you and home

so it is given as

v_1t + v_2t = d

4*t + 28*t = 3.2 km

t = \frac{3.2}{32} = 0.1 hr

so they will meet after t = 6 min

so from position time graph we can see that two will meet after t = 6 min where at this position two graphs will intersect


4 0
4 years ago
Neutron stars have been overtaken by what
Bogdan [553]
Gravity..? I don't know, but you need to clear up your question a bit. I'm sure it's gravity though.
6 0
3 years ago
A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the
insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}

Req =3.03Ω , R1 =12.18Ω

\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}

\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}

\frac{1}{R2}=0.2479

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0
4 years ago
How much force can a 2.5 kg sledge hammer excerpt on a nail if you can swing the hammer at 20 m/s and the hammer contacts the na
AleksAgata [21]

Answer:

625 N

Explanation:

The impulse given to the nail is equal to the change in momentum of the hammer:

I=F \Delta t=m \Delta v

where

F is the force exerted by the hammer

\Delta t=0.08 s is the time of contact

m=2.5 kg is the mass

\Delta v=20 m/s is the change in velocity of the hammer

Substituting the data and re-arranging the equation, we can find the force:

F=\frac{m \Delta v}{\Delta t}=\frac{(2.5 kg)(20 m/s)}{0.08 s}=625 N

4 0
4 years ago
A fidget spinner experiences a constant torque of 1.4 N.m. If the spinner is initially at rest, what is its angular momentum 2.0
motikmotik

Answer:

2.8N-msec

Explanation:

We have given torque \tau =1.4N-m

Initial time t_1=0\ sec

Final time t_2=2\ sec

There is relation between angular momentum and torque

that is \frac{dL}{dt}=\tau

dL=\tau dt

\int dL=\tau \oint_{t_1}^{t_2}dt

L=\tau (t_1-t_2)=1.4\times (2-0)=2.8N-msec

4 0
4 years ago
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