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3 years ago

72. An object travels at a constant velocity of 15 m/s for 5.0 seconds. How far does

Physics

1 answer:

Gennadij [26K]3 years ago

8 0

Answer: 125 m/s

Multiply 15 m/s by 5.0 minutes, because this will give you the product.

15 x 5.0 = 125.

Add the m/s sign to the 125. 125 m/s.

Answer = 125 m/s. Hope this helps!

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In Speed Study Number 1, we looked at two cars traveling the same distance at different speeds on city streets. Car "A" traveled

Ganezh [65]

Answer:

230.4 s

Explanation:

The speed of car A is

$v_A = 35 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car A is

$t_A = \frac{d}{v_A}=\frac{10 mi}{35 mi/h}=0.286 h$

The speed of car B is

$v_B = 45 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car B is

$t_B = \frac{d}{v_B}=\frac{10 mi}{45 mi/h}=0.222 h$

So the difference in time is

$\Delta t = t_A - t_B = 0.286 h -0.222 h=0.064 h$

Which corresponds to

$\Delta t = 0.064 h \cdot 3600 s/h = 230.4 s$

so car B arrived 230.4 s before car A.

5 0

2 years ago

In a nuclear reactor 1 g of mass is converted into energy. How much energy in Joules is produced?

KengaRu [80]

Answer:

3 x 10^5 J

Explanation:

mass of substance, m = 1 g = 0.001 kg

Velocity of light, c = 3 x 10^8 m/s

According to the Einstein mass energy equivalence, the energy associated with the mass is given by

E = m c^2

E = 0.001 x 3 x 10^8

E = 3 x 10^5 J

5 0

3 years ago

NEED HELP ASAP, ILL GIVE YOU BRAINLIEST IF CORRECT (30POINTS)

Vera_Pavlovna [14]

Answer:

the bar is the top and bottem. the nucleas in the middle and the Spiral arm is the last space

Explanation:

5 0

3 years ago

In some circumstances,science will answer questions about what is good and what is evil true or false?

bagirrra123 [75]

Your answer is false since science is a subject not a religion

4 0

3 years ago

Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8

sergejj [24]

Answer:

The hottest temperature is $T_2 = 39^o C$

Explanation:

From the question we are given

$T_1 = 98 F$

$T_2 = 39^oC$

$T_3 = 303 \ K$

Generally converting $T_3$ to Fahrenheit

$T_3' = (T_3 -273 ) * \frac{9}{5} + 32$

=> $T_3' = (303 -273 ) * \frac{9}{5} + 32$

=> $T_3' = 86 F$

Converting $T_2$ to Fahrenheit

$T_2' = T_2 * \frac{9}{5} + 32$

=> $T_2' = 39 * \frac{9}{5} + 32$

=> $T_2' =102.2 F$

Now comparing the temperature in Fahrenheit we see that $T_2$ is the hottest

3 0

2 years ago