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bija089 [108]
3 years ago
6

An object with a mass of 62.5 kg accelerates 12.3 m/s2 when an unknown force is applied to it. What is the amount of the force?

Physics
1 answer:
AlexFokin [52]3 years ago
4 0

Answer:768.75N

Explanation:

mass=62.5kg

acceleration=12.3m/s^2

Force=mass x acceleration

Force=62.5 x 12.3

Force=768.75N

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A horizontal spring with spring constant 950 N/m is attached to a wall. An athlete presses against the free end of the spring, c
OLga [1]

Answer:

66.5N

Explanation:

F = kx

Where F = force

K = spring constant

x = compression

Given

K = 950N/m

x = 7.0cm

F = ?

First convert the compression to meters .

7.0cm = 7.0 x 0.01

= 0.07 meters

Therefore

F = 950 x 0.07

= 66.5N

4 0
3 years ago
The speed of water flowing through a hose increases from 2.05 m/s to 31.4 m/s as it goes through the nozzle. What is the pressur
Nimfa-mama [501]

The pressure in the hose as the speed of water changes from 2.05 m/s to 31.4 m/s as it goes through the nozzle is 5.92 × 10⁵ N/m².

Given:

The flow of water through the hose initially, v₁ = 2.05 m/s

The flow of water through the hose initially, v₂ = 31.4 m/s

Calculation:

From Bernoulli's equation we have:

P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂

where P₁ is atmospheric pressure

           P₂ is the pressure in the hose

           ρ is the density of the fluid

           h₁ is the initial height

           h₂ is the final height

           v₁ is the initial velocity of the fluid

           v₂ is the final velocity of the fluid  and

           g is the acceleration due to gravity

Re-arranging the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (h₁-h₂)

Applying values in the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (0)

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [(31.4m/s)²-(2.05 m/s)²]

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [981.7575]

    = (1.01 × 10⁵ Pa)+ (4.91 × 10⁵ Pa)

    = 5.92 × 10⁵ Pa

    = 5.92 × 10⁵ N/m²

Therefore, the pressure in the hose is 5.92 × 10⁵ N/m².

Learn more about Bernoulli's equation here:

<u>brainly.com/question/9506577</u>

#SPJ4

6 0
1 year ago
What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are
nexus9112 [7]

Answer:

Density of 18.0-karat gold mixture is 15.58 g/cm^3.

Explanation:

A mixture of 18 parts gold, 5 parts silver, and 1 part copper.

Let mass of gold be 18x

Let the mass of silver be 5x

Let the mass of copper be 1x

The density of gold = 19.32g/cm^3

The density of silver = 10.1g/cm^3

The density of copper =8.8g/cm^3

Volume=\frac{Mass}{Density}

Volume of the gold in the mixture = V_1=\frac{18x}{19.32 g/cm^3}

Volume of the silver in the mixture = V_2=\frac{5x}{10.1 g/cm^3}

Volume of the copper in the mixture = V_3=\frac{1x}{8.8 g/cm^3}

Mass of the mixture = M = 18x+5x+1x =24x

Volume of the mixture = V_1+V_2+V_3

Density of the mixture:

\frac{M}{V_1+V_2+V_3}=15.58 g/cm^3

8 0
3 years ago
The rear window in a car is approximately a rectangle, 2.0 m wide and 0.65 m high. The inside rearview mirror is 0.50 m from the
Ymorist [56]

Answer:

0.43 m

Explanation:

Angle of incident and angle of reflection is same.

tan Θh = L' / x (eye)

L' = Length of the window

x (eye) = Distance of the mirror from the eye

tan Θh = L / (x (eye) + xw)

xw = Distance of the mirror from the window

L'/ x (eye) = L / ( x (eye) + xw)

L' = L*x (eye) / ( x (eye) + xw)

L' = (2*0.5) / (0.5 + 1.8)

L' = 0.43 m

5 0
3 years ago
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x= 5.0 cos
lakkis [162]

Answer:

a)   x = 4.33 m ,   b)  w = 2 rad / s ,  f = 0.318 Hz ,  c) a = - 17.31 cm / s²,  

d) T =  3.15 s,  e)  A = 5.0 cm

Explanation:

In this exercise on simple harmonic motion we are given the expression for motion

          x = 5 cos (2t + π / 6)

they ask us for t = 0

a) the position of the particle

      x = 5 cos (π / 6)

      x = 4.33 m

remember angles are in radians

 

b) The general form of the equation is

          x = A cos (w t + Ф)

when comparing the two equations

         w = 2 rad / s

angular velocity and frequency are related

          w = 2π f

           f = w / 2π

           f = 2 / 2pi

           f = 0.318 Hz

c) the acceleration is defined by

      a == d²x / dt²

      a = - A w² cos (wt + Ф)

for t = 0 ,  we substitute

      a = - 5,0 2² cos (π / 6)

      a = - 17.31 cm / s²

d) El period is

          T = 1/f

         T= 1/0.318

         T =  3.15 s

e) the amplitude

        A = 5.0 cm

3 0
3 years ago
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