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Charra [1.4K]
3 years ago
12

A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang

le made by the string when the mass comes to momentary rest
​
Physics
1 answer:
Dafna11 [192]3 years ago
8 0

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) E=\frac{1}{2}mv^2+mgh

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) E=\frac{1}{2}mv^2

The energy when the mass comes to a stop will be:

3) E=mgh

Setting equations 2 and 3 equal and solving for height h will give:

4) h=\frac{v^2}{2g}

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) cos\phi=\frac{l-h}{l}

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) \phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})

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riadik2000 [5.3K]

Answer:

The horizontal component of the truck's velocity is: 23.70 m/s

The vertical component of the truck's velocity is: 3.13 m/s

Explanation:

You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.

The identities are:

Cosα= \frac{CA}{H}

Senα= \frac{CO}{H}

Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus

The horizontal component of the truck's velocity is:

Let Vx represent it.

In this case, CA=Vx, H=24 and α=7.5 degrees

Vx=(24)Cos(7.5)

Vx=23.79 m/s

The vertical component of the truck's velocity is:

Let Vy represent it.

In this case, CO=Vy, H=24 and α=7.5 degrees

Vy=(24)Sen(7.5)

Vy=3.13 m/s

3 0
3 years ago
#2 A car accelerates from rest at 4 m/s^2. What is the velocity of the car after 4 second?
solmaris [256]
You use the equation Velocity = Acceleration X Time. 4x4=16m/s.


The car travels 18m in 3 seconds.
3 0
3 years ago
Maria is comparing data from an investigation about the melting point of ice. Based on her research, she knows ice
erica [24]

Answer:

The answer Is set 4.

Explanation

8 0
3 years ago
Read 2 more answers
An electric furnace runs 13 hours a day to heat a house during January (31 days). The heating element has a resistance of 7.2 an
liq [111]

Answer:

cost of running the furnace during January is $5619.62

Explanation:

given data

runs a day = 13 hours

January days = 31 days

resistance = 7.2 ohm

current = 16.7 A

cost of electricity = $0.10/kWh

to find out

cost of running the furnace during January

solution

first we get her power consumed by furnace that is

Power consumed = \frac{I^2}{R}  ........1

put here value we get

Power consumed = \frac{16.7^2}{7.2}

Power consumed = 38.7347 W

and

Power consumed by furnace in one hour is

Power consumed by furnace in one hour is = Power consumed × 3600

Power consumed by furnace in one hour is = 38.7347 × 3600  

Power consumed by furnace in one hour is 139.445kWh

and

Power consumed by furnace in the month of January is

Power consumed by furnace in the month of January = 139.445kWh × 13 hours × 31 days

Power consumed by furnace in the month of January = 56196.335 kWh

so

cost of running the furnace during January is = $0.10/kWh × 56196.335 kWh

cost of running the furnace during January is $5619.62

4 0
3 years ago
Read 2 more answers
A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.
hichkok12 [17]

Answer:

a) \Delta U_g=-5.3kJ

b) K=0.27kJ

c) F_f=0.45kN

Explanation:

the gravitational potential energy is given by:

U_g=m.g.h\\

\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ

The kinetic energy is given by:

K=\frac{1}{2}m.v^2\\

the initial kinetic energy is zero because the motion started from rest, so:

K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ

applying the conservation of energy theorem:

U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ

The work done by the friction force is given by:

W_f=F_f.h.cos(\theta)\\

the angle of the force is 180 degrees because it's against the movement:

F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN

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