How many grams of ice would have to melt to lower the temperature of 352 ml of water from 25?

An automobile traveling along a straight road increases its speed from 72 ft/s to 84 ft/s in 180 ft. if the acceleration is cons

Nikolay [14]

The equation that would allow us to calculate for the acceleration given the distance is written below,

a = (Vf² - Vo²) / 2d

where a is the acceleration, Vf is the final velocity, Vo is the initial velocity, and d is distance.

Substituting the known values,
a = ((84 ft/s)² - (72 ft/s)²) / 2(180 ft) = 5.2 ft/s²

Then, the equation that would relate the initial velocity, distance, acceleration and time is calculated through the equation,

d = Vot + 0.5at²

Substituting the known values,
180 = 72(t) + 0.5(5.2)(t²)

The value of t from the equation is 2.3 s

ANSWER: 2.3 s

5 0

4 years ago

It is the thermal energy transferred from a hot object to a cold object.

kakasveta [241]

It is the thermal energy transferred from a hot object to a cool object.

The following statement listed above is correct.

Answer: True.

6 0

4 years ago

A 15.0 Kg object is moved from a height of 7.00 m above thefloor to a height of 13.0 m above the floor. What is the change ingra

Pie

To solve this problem we will apply the concepts related to potential gravitational energy. This is defined as the product between mass, acceleration and change in height and can be expressed as,

$\Delta PE = mg \Delta h$

Here,

m = Mass

g = Gravitational acceleration

$\Delta h$ = Height

Replacing with our values we have,

$\Delta PE = (15kg)(9.81m/s^2)(13m-7m)$

$\Delta PE = 882.9J \approx 883J$

Therefore the change in gravitational potential energy is 883J.

4 0

3 years ago

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket

NARA [144]

The constant acceleration of a rocket launched upward, calculated knowing that the time it takes for a bolt that falls off the side of the rocker was 6.30 seconds, is 5.68 m/s².

When the rocket is launched straight up with constant acceleration, the acceleration of the rocket is given by:

$v_{f_{r}} = v_{i_{r}} + at$