If you move 50 meters in 10 seconds, what is your speed

A cyclist is riding his bike up a mountain trail. When he starts up the trail, he is going 8 m/s. As the trail gets steeper, he

shusha [124]

Answer:

a) a = - 0.0833 m / s², b) t = 4.4 s

Explanation:

a) this is a kinematics exercise where the acceleration is along the inclined plane

v = v₀ - a t

a = v₀ - v / t

a = 3 - 8/60

a = - 0.0833 m / s²

b) in this case the final velocity is zero

v = v₀ - a t

0 = v₀ - at

t = v₀ / a

t = 28 / 6.4

t = 4.375 s

t = 4.4 s

3 0

3 years ago

A car is cruising at a steady speed of 35 mph. Suddenly, a cuddly puppy runs out into the road. The driver takes 1.7 seconds to

Schach [20]

Answer:

The distance traveled is 0.037 mi

Explanation:

The equation for the position and velocity of an accelerated object is:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a * t

where

x = position at time t

x0 = initial position

t = time

a = acceleration

v0 = initial velocity

If the velocity is constant, then a = 0 and the position will be:

x = x0 + v * t where "v" is the velocity

First, let´s find the distance traveled until the driver push the brake:

The speed is constant. Then:

x = x0 + v * t (considering the origin of the reference system to be located at the point at which the driver sees the puppy, x0 = 0)

x = 35 mi/h (1 h / 3600 s) * 1.7 s = 0.017 mi

Then, the drivers moves with constant acceleration until the car stops (v = 0)

From the equation for velocity:

v = v0 + a * t

Since v = 0, we can obtain the acceleration of the car until it stops. With that acceleration, we can calculate how much distance the car moves before it stops.

0 = v0 + a * t

-v0 / t = a

-35 mi/h (1 h / 3600s) / 4.0 s = a

a = -2.4 x 10⁻³ mi/s²

The distance traveled will be:

x = x0 + v0 * t + 1/2 * a * t²

Now x0 will be the distance traveled before the driver slows down.

x = 0.017 mi + 35 mi/h (1 h / 3600s) * 4 s + 1/2 * ( -2.4 x 10⁻³ mi/s²) * (4s)²

x = 0.037 mi

6 0

3 years ago

An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Alexeev081 [22]

Answer:

a) $c=1822.3214\ J.kg^{-1}.K^{-1}$

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

mass of aluminium, $m_a=0.1\ kg$

mass of water, $m_w=0.25\ kg$

initial temperature of the system, $T_i=10^{\circ}C$

mass of copper block, $m_c=0.1\ kg$

temperature of copper block, $T_c=50^{\circ}C$

mass of the other block, $m=0.07\ kg$

temperature of the other block, $T=100^{\circ}C$

final equilibrium temperature, $T_f=20^{\circ}C$

We have,

specific heat of aluminium, $c_a=910\ J.kg^{-1}.K^{-1}$

specific heat of copper, $c_c=390\ J.kg^{-1}.K^{-1}$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

$Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)$

$Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)$

$Q_{in}=11375\ J$

The heat released by the blocks when dipped into water:

$Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)$

where

$c=$ specific heat of the unknown material

For the conservation of energy : $Q_{in}=Q_{out}$

so,

$11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)$

$c=1822.3214\ J.kg^{-1}.K^{-1}$

b)

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

7 0

3 years ago

What is the primary determinant of the voltage developed by a battery?

Fittoniya [83]

For the answer to the question above asking what is the primary determinant of the voltage developed by a battery?the answer is that the <span>the nature of the materials in the reaction that is the primary determinant of the voltage from a battery.</span>

5 0

3 years ago

A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.

konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

Mathematically:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$