Ask question

Ask question

All categories

2 years ago

8

A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus

e the parachute is going down.) It comes to rest at 0 meters/second after 30 seconds. What's the average acceleration of the shuttle during this time span?

1 answer:

dexar [7]2 years ago

6 0

Answer:

$a=5m/s^2$

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

$v_{1}=150m/s$

and the final velocity is :

$v_{2}=0 m/s$

so, the change in velocity is:

$\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) = 150 m/s$

and the change in time , according to the problem:

$\Delta t=30s$

So, the aceleration is:

$a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2$

You might be interested in

The total negative charge on the electrons in 1kg of helium (atomic number 2, molar mass 4) is____________.

tekilochka [14]

Answer:

Explanation:

$n = \frac{m}{M}$

$n = \frac{1000}{4}$

= 250 moles.

N = n×6.02× $10^{23}$

= 1.505× $10^{26}$

Total charge = (1.505× $10^{26}$ ) × (1.6× $10^{-19}$ )

= 2.4× $10^{7}$ C.

4 0

2 years ago

Read 2 more answers

The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy o

MrMuchimi

Answer:

The total kinetic energy of both particles is $2.43\times10^{-13}$

Explanation:

Given that,

Kinetic energy of nucleus $K.E= 1.24\times10^{-13}\ J$

Kinetic energy of proton $K.E= 2.47\times10^{-13}\ J$

Radius of proton $r= 0.9\times10^{-15}\ m$

We need to calculate the final potential energy

Using formula of final potential energy

$U=\dfrac{kq^2}{r}$

Put the value into the formula

$U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times0.9\times10^{-15}}$

$U_{f}=1.28\times10^{-13}\ J$

We need to calculate the initial energy of both the particles

Using formula of energy

$E_{i}=(K.E_{n}+K.E_{p})+U_{i}$

$E_{i}=1.24\times10^{-13}+2.47\times10^{-13}+0$

$E_{i}=3.71\times10^{-13}\ J$

We need to calculate the total kinetic energy of both particles

Using conservation of energy

$E_{i}=E_{f}$

$E_{i}=K.E_{f}+U_{f}$

$3.71\times10^{-13}=K.E_{f}+1.28\times10^{-13}$

$K.E_{f}=3.71\times10^{-13}-1.28\times10^{-13}$

$K.E_{f}=2.43\times10^{-13}$

Hence, The total kinetic energy of both particles is $2.43\times10^{-13}$

3 0

1 year ago

The temperature of a heat engine is 500k some of the heat generated by the engine flows to the surroundings which are at a temp

Ierofanga [76]

1-125/500)x100=efficiency

1-1/4)x100 =75pc

3 0

1 year ago

A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra

skelet666 [1.2K]

Answer:

So the acceleration of the child will be $8.05m/sec^2$

Explanation:

We have given angular speed of the child $\omega =1.25rad/sec$

Radius r = 4.65 m

Angular acceleration $\alpha =0.745rad/sec^2$

We know that linear velocity is given by $v=\omega r=1.25\times 4.65=5.815m/sec$

We know that radial acceleration is given by $a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2$

Tangential acceleration is given by

$a_t=\alpha r=0.745\times 4.65=3.464m/sec^$

So total acceleration will be $a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2$

7 0

2 years ago

A student produces a power of p = 0.87 kw while pushing a block of mass m = 75 kg on an inclined surface making an angle of Î¸ =

Paul [167]

When block is pushed upwards along the inclined plane

the net force applied on the block will be given as

$F_{net} = mg sin\theta + \mu_k mg cos\theta$

here we know that

m = 75 kg

$\theta = 8.5 degree$

$\mu_k = 0.16$

now plug in all values into this

$F_{net} = 75\times 9.8 sin8.5 + 0.16 \times 75\times 9.8 cos8.5$

$F = 225 N$

now for finding the power is given as

$P = Fv$

$0.87 \times 10^3 = 225 \time v$

$v = \frac{870}{225} = 3.87 m/s$

6 0

1 year ago