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Vesnalui [34]
3 years ago
15

Guys pa help...Sana po yong totoo...wag napo sumagot Kung do nman alam​

Physics
1 answer:
rodikova [14]3 years ago
6 0
Gdhootiytuyin bilhyffjoueriiggyi
You might be interested in
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
A 115 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice.
Hoochie [10]

Answer:

(a) Ff = 0.128 N

(b μk = 0.1135

Explanation:

kinematic analysis

Because the hockey puck  moves with uniformly accelerated movement we apply the following formulas:

vf=v₀+a*t Formula (1)

d= v₀t+ (1/2)*a*t² Formula (2)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s

Calculation of the acceleration of the  hockey puck

We apply the Formula (1)

vf=v₀+a*t      v₀=5.8 m/s ,  vf=0

0=5.8+a*t

-5.8 = a*t

a= -5.8/t   Equation (1)

We replace a= -5.8/t in the Formula (2)

d= v₀*t+ (1/2)*a*t²   ,  d=15.1 m ,  v₀=5.8 m/s

15.1 = 5.8*t+ (1/2)*(-5.8/t)*t²  

15.1= 5.8*t-2.9*t

15.1= 2.9*t

t = 15.1 / 2.9

t= 5.2 s

We replace t= 5.2 s in the equation (1)

a= -5.8/5.2

a= -1.115 m/s²

(a) Calculation of the  frictional force (Ff)

We apply Newton's second law

∑F = m*a    Formula (3)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Look at the free body diagram of the  hockey puck in the attached graphic

∑Fx = m*a     m= 115g * 10⁻³ Kg/g = 0.115g    ,  a= -1.12 m/s²

-Ff = 0.115*(-1.115)  We multiply by (-1 ) on both sides of the equation

Ff = 0.128 N

(b) Calculation of the coefficient of friction (μk)

N: Normal Force (N)

W=m*g= 0.115*9.8= 1.127 N : hockey puck  Weight

g: acceleration due to gravity =9.8 m/s²

∑Fy = 0

N-W=0

N = W

N =  1.127 N

μk = Ff/N

μk = 0.128/1.127

μk = 0.1135

8 0
3 years ago
Using a mass of 1.8 g and the volume displaced by the sample, calculate the sample's density. A) 0.17 g/mL B) 0.35 g/mL C) 0.60
fomenos
The answer to that probably would be C excuse me if I am wrong.
8 0
3 years ago
Read 2 more answers
This chart lists four examples of two objects that are in contact.
Oxana [17]

Before coming into conclusion first we have to understand the direction of heat flow.

Heat is the transferred thermal energy from one body to another body due to the temperature difference just like water flows from higher level to lower level.

Whenever two bodies having different temperature come closer to each other heat will flow from hotter body to cooler one if no external work is done. The heat flow may be through any of the ways i.e conduction,radiation or convection. Hence temperature difference is the parameter which gives the direction of heat flow.

The temperature is also considered as a measure of average kinetic energy of the substance.The thermal energy does not give the direction heat flow. Heat may  flow from the body  having low thermal energy but at higher temperature to the body having higher thermal energy but at low temperature. The reverse does not happen naturally .

In example 1 there is fire and air. Obviously fire is at high temperature and air at low temperature.So heat will flow from object 1 to object 2.

In example 2 there is a metal at 80 degree Celsius and another metal at 12 degree Celsius .So heat will flow from object 1 to object 2

In example 3 we have cooler ocean and warm air. So the heat will flow from object 2 to object 1.

In example 4 we have a tool with high thermal energy and a material with little thermal energy.We already know that thermal energy can not determine the direction of heat flow. Here the temperature of each substance is not given.The kinetic energy is  part of thermal energy.So there is the chance of higher kinetic energy of the tool for having higher thermal energy .At that time the heat will flow object 1 to object 2.Otherwise the reverse will occur. So it is a special case.

As per the question only option 4 is correct which tells that heat will flow from object 1 to object 2 in examples 1,2,4, and heat will flow from object 2 to 1 in example 3. Other options violate the fundamental law of thermodynamics.


7 0
3 years ago
A spring has a spring constant of 90N/m.How much potential energy does it store when stretched by 2 cm?
il63 [147K]

Answer:

The potential energy stored in the spring is 0.018 J.

Explanation:

Given;

spring constant, k = 90 N/m

extension of the spring, x = 2 cm = 0.02 m

The potential energy stored in the spring is calculated as;

U = ¹/₂kx²

where;

U is the potential energy stored in the spring

Substitute the given values in the equation above;

U = ¹/₂  x  90 N/m  x  (0.02 m)²

U = 0.018 J

Therefore, the  potential energy stored in the spring is 0.018 J.

5 0
3 years ago
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