Ask question

Ask question

All categories

2 years ago

6

A rocket achieves a lift-off velocity of 500.0 m/s from rest in 30.0 seconds. Calculate the average acceleration of the rocket.

1 answer:

Anettt [7]2 years ago

6 0

Answer:

The average acceleration is 16.6 m/s² ⇒ 1st answer

Explanation:

A rocket achieves a lift-off velocity of 500.0 m/s from rest in

30.0 seconds

The given is:

→ The initial velocity = 0

→ The final velocity = 500 meters per seconds

→ The time is 30 seconds

Acceleration is the rate of change of velocity of the rocket

→ $a=\frac{v-u}{t}$

where a is the acceleration, v is the final velocity, u is the initial velocity

and t is the time

→ u = 0 , v = 500 m/s , t = 30 s

Substitute these values in the rule

→ $a=\frac{500-0}{30}=\frac{500}{30}=16.6$ m/s²

<em>The average acceleration is 16.6 m/s²</em>

You might be interested in

Suppose I have a vector that is 7 units long and that makes an angle of +30 degrees from the positive x-axis. I want to add to t

Vinil7 [7]

Answer:

sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

Explanation:

We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis

So x component of the vector $=7cos30^{\circ}=7\times 0.866=6.06$

y component of the vector $=7sin30^{\circ}=7\times 0.5=3.5$

So vector will be 6.06i+3.5j

Now other vector of length of 7 units and makes an angle of 120° with positive x-axis

So x component of vector $=7cos120^{\circ}=7\times -0.5=-3.5i$

y component of the vector $=7sin120^{\circ}=7\times 0.866=6.06j$

Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

5 0

3 years ago

39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th

tatuchka [14]

<u>Answer:</u> The final temperature of the solution is $80.14^oC$

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $\text{heat}_{absorbed}=\text{heat}_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 39 g

$m_2$ = mass of coffee = 166 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $83^oC$

$c_1$ = specific heat of aluminium = $0.904J/g^oC$

$c_2$ = specific heat of coffee= $4.1801J/g^oC$

Putting all the values in equation 1, we get:

$39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]$

$T_{final}=80.14^oC$

Hence, the final temperature of the solution is $80.14^oC$

4 0

3 years ago

Waves can carry either ____ or __________ energy?

Oksanka [162]

Mechanical or Electromagnetic

3 0

2 years ago

Read 2 more answers

When a pitcher is throwing a ball, is he applying positive or negative work to the ball?

Romashka [77]

A.Positive as work=force×distance (assuming that distance is always a constant) the work will always be positive as the force is always positive(this is because force=mass×acceleration where the acceleration is always positive unless the mass is being slowed down)

5 0

3 years ago

Read 2 more answers

A plant worker accidentally breathes some stored gaseous tritium, a beta emitter with maximum particle energy of 0.0186 MeV. The

andrey2020 [161]

Answer:

400 milli-rems

Solution:

As per the question;

Maximum energy of particle, $E_{m} = 0.0186\ MeV$

Weight, w = 1 kg

Energy absorbed, E = $4\times 10^{- 3}\ J$

Now,

Equivalent dose is given by:

$D_{eq} = \frac{E}{w} =\frac{4\times 10^{- 3}}{1} = 4\times 10^{- 3}\ J/kg$

1 Gy = 1 J/kg

Also,

1 Gy = $1\times 10^{5}\ milli-rems$

Therefore,

Dose equivalent in milli-rems is given by:

$D_{eq} = 4\times 10^{- 3}\times 1\times 10^{5} = 400\ milli-rems$

8 0

3 years ago