How long does it take a car traveling at 50 mph to travel 75 miles? Use one of the following to find the answer.

An electric motor rotating a workshop grinding wheel at 1.06 102 rev/min is switched off. Assume the wheel has a constant negati

kvasek [131]

Answer:

t = 106π / 30*2.1

Explanation:

$w_{i} = 1.06*10^{2}$ => 106

=> 106 x 2π/60

=> 106/30π

∝ = -2.1 rad/sec²

$w_{f}$ => 0

$w_{f}$ = $w_{i}$ + ∝t

∴ ( $w_{f}$ - $w_{i}$ ) / ∝ = t

t = 106π / 30*2.1

6 0

3 years ago

What type of reaction feels cold to the touch? A. Endothermic, because energy is being released into the surroundings B. Endothe

Crazy boy [7]

Answer:

D. Exothermic, because energy is being absorbed from the surroundings

Explanation:

This is true about the Exothemic reaction due to the fact that, the reaction occurs outside the body. During this reaction, the energy being absorbed <em>from the surrounding environment will hit the body surface thereby creating the coldness due to the heat given out from the body being minimal.</em>

8 0

2 years ago

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The LR5 is the specialist submarine for underwater rescue. The average density of seawater is 1028 kg/ m3.

sladkih [1.3K]

Answer:

P = 7196 [kPa]

Explanation:

We can solve this problem using the expression that defines the pressure depending on the height of water column.

P = dens*g*h

where:

dens = 1028 [kg/m^3]

g = 10 [m/s^2]

h = 700 [m]

Therefore:

P = 1028*10*700

P = 7196000 [Pa]

P = 7196 [kPa]

5 0

3 years ago

The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared

velikii [3]

Answer:

$\displaystyle a(5)=-32$

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

$\displaystyle v(t)=\frac{df}{dt}$

And the acceleration is

$\displaystyle a(t)=\frac{dv}{dt}$

Or equivalently

$\displaystyle a(t)=\frac{d^2f}{d^2t}$

The given height of a projectile is

$f(t)=-16t^2 +238t+3$

Let's compute the speed

$\displaystyle v(t)=-32t+238$

And the acceleration

$\displaystyle a(t)=-32$

It's a constant value regardless of the time t, thus

$\boxed{\displaystyle a(5)=-32}$

3 0

3 years ago

A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .

rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

$-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0$

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

$v=u+at$

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

$v = 15.5 +(-9.8)(2.64)=-10.4 m/s$

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

$-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0$

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

5 0

3 years ago

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