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3 years ago

How is air resistance similar to gravity? give me two reasons

Physics

1 answer:

QveST [7]3 years ago

8 0

Explanation:

I think this would help you. Read this and make your own answer ok.

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If a power lifter raises a 1000N weight a distance of 2 meters in 0.5 seconds, what is his power out put?

Irina18 [472]

Answer:

P = 4000 [W]

Explanation:

In order to solve this problem, we must first determine the work, which is defined as the product of force by distance.

W = F*d

where:

W = work [J] (units in Joules)

F = force = 1000[N]

d = distance = 2 [m]

W = 1000*2

W = 2000 [J]

And power is defined as the relationship between work and the time in which the work is done.

P = W/t

P = power [W] (units of watts)

t = time = 0.5 [s]

P = 2000/0.5

P = 4000 [W]

8 0

3 years ago

The temperature of a lead fishing weight rises from 26∘c to 38∘c as it absorbs 11.3 j of heat. what is the mass of the fishing w

Nimfa-mama [501]

Given:
ΔT = 38 - 26 = 12°C, temperature change
Q = 11.3 J, heat input
c = 0.128 J/(g-°C), specific heat of lead

Let m = the mass of the lead.
Then
Q = m*c*ΔT
(m g)*(0.128 J/(g-°C))*(12 °C) = 11.3 J
1.536m = 11.3
m = 7.357 g

Answer: 7.36 g (2 sig. figs)

8 0

3 years ago

In Fig., block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.50 m and then collides with stationary

Evgesh-ka [11]

Answer:

Explanation:

Given

initial height $h=2.5 m$

$m_2=2m_1$

coefficient of static friction $\mu =0.5$

When collision is elastic respective velocities after collision is

$v_1=\frac{u_1(m_1-m_2)+2m_2u_2}{m_1+m_2}$

$v_2=\frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2}$

where $u_1$ and $u_2$ =initial velocities of object

$v_1$ and $v_2$ final velocities of object

$u_1=\sqrt{2\times 9.8\times 2.5}$

$u_1=7 m/s$

$v_2=\frac{0+2m_1\times 7}{m_1+2m_1}$

$v_2=\frac{14}{3} m/s$

using $v^2-u^2=2 as$

$0-(4.67)^2=2\times (-0.5\times 9.8)\times s$

$s=2.22\ m$

(b)Completely Inelastic

In Completely Inelastic objects stick with each other

$m_1u_1=(m_1+m_2)v$

$v=\frac{u_1}{3}=\frac{7}{3} m/s$

using $v^2-u^2=2 as$

$0-(2.33)^2=2\times (-0.5\times 9.8)\times s$

$s=0.55\ m$

7 0

4 years ago

A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is

Dennis_Churaev [7]

Answer:

F = 789 Newton

Explanation:

Given that,

Speed of the car, v = 10 m/s

Radius of circular path, r = 30 m

Mass of the passenger, m = 60 kg

To find :

The normal force exerted by the seat of the car when the it is at the bottom of the depression.

Solution,

Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.

$N=mg+\dfrac{mv^2}{r}$

$N=m(g+\dfrac{v^2}{r})$

$N=60\times (9.81+\dfrac{(10)^2}{30})$

N = 788.6 Newton

N = 789 Newton

So, the normal force exerted by the seat of the car is 789 Newton.

6 0

3 years ago

How long does it take a person to skate the width of a hockey rink (85 feet) at a constant speed of 15 feet per second?

Zepler [3.9K]

S(travel distance)=85 ft
v (velocity)=15 ft/s
-----------------------------------
t (time)=?

Calculate the time with the formula for the velocity:
v=S/t
t=S/v
t=85 ft/(15 ft/s)
t=5.666s

3 0

3 years ago