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2 years ago

11

A car is driving along a road at a speed of 50 km/hr. How far will the car travel in 5 hours?

1 answer:

BARSIC [14]2 years ago

4 0

Answer:

I think it's 250

Explanation:

If the car is traveling 50 km/hr that means every hour, the car drives 50 km. So if you want to know how far it will go in 5 hours you do 50x5.

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Can someone help me with this. I'm not really sure if the right answer is c.

taurus [48]

It might be c I'm not sure either. Hope I helped

6 0

3 years ago

An air conditioner has a power rating of 3000 W and is used daily. Its electrical

Anarel [89]

Answer:

5.714 hours / day

Explanation:

<u>Calculate the hours used in that week </u>

120000 / 3000 = 120 / 3 = 40 hours a week

<u>Calculate the amount it is used in one day</u>

40 / 7 = 5.71428571 hours or 5.714 hours/day

5 0

2 years ago

Read 2 more answers

A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m

elena-14-01-66 [18.8K]

Answer:

<u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

$f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration$

<u>2. Formulae:</u>

$f=\dfrac{number\text{ }of\text{ }revolutions}{time}$

$\omega=2\pi f$

$a_c=\omega^2 r$

$F_c=m\times a_c$

<u>3. Solution (calculations)</u>

$f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}$

$\omega=2\pi\times0.\overline{60}\approx 3.808rad/s$

$a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2$

$F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N$

3 0

3 years ago

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago

Radioactive isotopes can be used to find the age of rocks, fossils, or other artifacts. Carbon-14 has a half-life of 5,730 years

bazaltina [42]

Answer:

1/8 = (1/2)^3

This implies the sample has decayed for 3 half lives

3 * 5730 yrs = 17,200 years

8 0

2 years ago