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3 years ago

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

2 years 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