All categories

3 years ago

The initial speed of a body is 3.28 m/s. What is its speed after 2.32 s if it accelerates

1 answer:

6 0

Final velocity = initial velocity + acceleration * time

v = u + at

v = 3.28 + 2.32 * 2.08

v = 3.28 + 4.83

<u>v = 8.11 m/s</u>

You might be interested in

Physics question, answer completely with work

Alenkasestr [34]

The force is 2.0 N east

Explanation:

The impulse exerted by a force is defined as the product between the force itself and the time interval during which the force is applied. Mathematically, it is equal to the change in momentum experienced by the object on which the force is acting:

$I=F\Delta t = \Delta p$

Where

I is the impulse

F is the force

$Delta t$ is the time interval during which the force is applied

$\Delta p$ is the change in momentum

In this problem,

$\Delta t = 3.0 s$ is the time interval

$I=6.0 N\cdot s$ (east) is the impulse

Therefore, the magnitude of the force is

$F=\frac{I}{\Delta t}=\frac{6.0}{3.0}=2.0 N$

And the direction is the same as the impulse (east).

Learn more about impulse and change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

7 0

2 years ago

If a green light shines on a red toy, the toy will look:

disa [49]

Not 100% sire but I think it'd be Yellow since we see red and green light together as Yellow

4 0

2 years ago

A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and

snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

$\dfrac{dp}{dt}=F$

$\dfrac{dP}{dt}=\dfrac{mv^2}{r}$

Put the value into the formula

$\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}$

$\dfrac{dP}{dt}=794.11\ N$

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0

3 years ago

Years ago, a block of ice with a mass of about 20.0 kg was used daily in a home icebox. The temperature of the ice was 0.0°C whe

m_a_m_a [10]

Answer: The ice absorb 6671.1 kJ of thermal energy.

Explanation:

The conversions involved in this process are :

$0.00^0C=273K$

$:H_2O(s)(273K)\rightarrow H_2O(l)(273K)$

Now we have to calculate the enthalpy change.

$\Delta H=n\times \Delta H_{fusion}$

where,

$\Delta H$ = enthalpy change = ?

m = mass of ice = 20.0 kg = $20.0\times 10^3g$ (1kg=1000g)

n = number of moles of ice= $\frac{\text{Mass of ice}}{\text{Molar mass of water}}=\frac{20.0\times 10^3g}{18g/mole}=1.11\times 10^3mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

Now put all the given values in the above expression, we get

$\Delta H=1.11\times 10^3mole\times 6010J/mole$

$\Delta H=6671100J=6671.1kJ$ (1 kJ = 1000 J)

Therefore, the enthalpy change is, 6671.1 kJ

6 0

2 years ago

Help . . . . . . . . . . . . . . . . . . . .

Zigmanuir [339]

Explanation:

Transmission

Reflection

Absorption

Refraction

Diffraction

4 0

2 years ago