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Elena-2011 [213]

3 years ago

14

an arrow is launched straight up from the ground with an initial velocity of 23.4 m/s. how long until it reaches its highest poi

nt?

1 answer:

AnnyKZ [126]3 years ago

4 0

Explanation:

Given that,

The initial velocity of an arrow is 23.4 m/s

We need to find the time taken by it to reach its maximum height. Let the maximum height is h.

Using third equation of motion to find height first.

$v^2-u^2=2ah$

At maximum height, v = 0

a = -g

$u^2=2gh\\\\h=\dfrac{u^2}{2g}\\\\h=\dfrac{(23.4)^2}{2\times 9.8}\\\\h=27.93\ m$

Let t is the time to reach its maximum height. So, it can be calculated using second equation motion as follows :

$27.93=23.4t-\dfrac{1}{2}\cdot9.8\cdot t^{2}\\\\t=2.35\ s$

Hence, it will take 2.35 seconds to reach its maximum height.

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a spring is stretched 2 cm when a mass of 40 grams is hung from it what is the spring constant of the spring

Aliun [14]

F = ma = -kx
m = 0.04 kg
x = -0.02m
a = 9.81m/s²

k = ma/x = 19.6 N/m

5 0

3 years ago

Read 2 more answers

A circular-motion addict of mass 83.0 kg rides a Ferris wheel around in a vertical circle of radius 13.0 m at a constant speed o

Svetllana [295]

Answer:

(A) Time period T = 6.28 SEC

(B) At highest point fore is - 575.83 N

(B) At lowest point force is 1050.97 N

Explanation:

We have given that mass m = 83 kg

Radius r = 13 m

Speed v = 6.10 m/sec

(A) Time period of the motion is given by $T=\frac{2\pi r}{v}=\frac{2\times 3.14\times 13}{6.10}=6.28sec$

(b) Net force is given by $F_{NET}=\frac{mv^2}{r}=\frac{83\times 6.10^2}{13}=237.571N$

Force due to gravity $F_{gravity}=mg=83\times 9.8=813.4N$

At highest point $F_{NORMAL}=F_{NET}-F_{GRAVITY}=237.57-813.4=-575.83$

(B) At lowest point $F_{NORMAL}=F_{NET}+F_{GRAVITY}=237.57+813.4=1050.97N$

5 0

3 years ago

When an aluminum bar is connected between a hot reservoir at 720 K and a cold reservoir at 358 K, 3.00 kJ of energy is transferr

disa [49]

Answer:

a. -4.166 J/K

b. 8.37 J/K

c. 4.21 J/K

d. entropy always increases.

Explanation:

Given :

Temperature at hot reservoir , $T_h$ = 720 K

Temperature at cold reservoir , $T_c$ = 358 K

Transfer of heat, dQ = 3.00 kJ = 3000 J

(a). In the hot reservoir, the change of entropy is given by:

$dS_h= -\frac{dQ}{t_h}$ (the negative sign shows the loss of heat)

$dS_h= -\frac{3000}{720}$

= -4.166 J/K

(b) In the cold reservoir, the change of entropy is given by:

$dS_c= \frac{dQ}{t_c}$

$dS_c= \frac{3000}{358}$

= 8.37 J/K

(c). The entropy change in the universe is given by:

$dS=dS_h+dS_c$

= -4.16+8.37

= 4.21 J/K

(d). According to the concept of entropy, the entropy of the universe is always increasing and never decreasing for an irreversible process. If the entropy of universe decreases, it violates the laws of thermodynamics. Hence, in part (c), the result have to be positive.

5 0

3 years ago

A truck brakes from 15 m/s to 2 m/s in 15 seconds. What is its acceleration?

iris [78.8K]

If we take v=u+at and our given values are:

v=final velocity 20m/s

u=initial velocity 15m/s

a=acceleration

t=time 5sec

we plug those into our equation:

20m/s=15m/s+a(5s)

now we just rearrange to solve for acceleration.

(5m/s)/(5s)=a

a=1m/s^2

hope this helps!
PLEASE MARK BRAINLEIST

3 0

3 years ago

A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center.

My name is Ann [436]

Answer:

angular momentum, $L=1.4\times 10^4\ kg-m^2/s$

Explanation:

Given that,

Mass of the woman, m = 50 kg

Angular velocity of the disk, $\omega=0.8\ rev/s=5.02\ rad/s$

Mass of the disk, m' = 2670 kg

Radius of the disk, R = 4 m

We need to find the magnitude of the total angular momentum of the woman–disk system. The moment of inertia of the system is equal to the sum of moment of inertia of women and the moment off inertia of the disk.

$I=mR^2+\dfrac{1}{2}m'R^2\\\\I=R^2(m+\dfrac{1}{2}m')\\\\I=(4)^2\times (50+\dfrac{1}{2}\times 270)\\\\I=2960\ kg-m^2$

The angular momentum is given by :

$L=I\omega\\\\L=2960\times 5.02\\\\L=14859.2\ kg-m^2/s$

or

$L=1.4\times 10^4\ kg-m^2/s$

So, the magnitude of the total angular momentum of the woman–disk system is $1.4\times 10^4\ kg-m^2/s$ . Hence, this is the required solution.

7 0

3 years ago