Someone can help me pls Is physical science class

a car accelerates uniformly from rest to a speed of 65 km/h (18 m/s) in 12s. Find the distance the car travels during this time?

Vinil7 [7]

The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it. Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.

12 seconds at an average speed of 9 m/s ==> (12 x 9) = 108 meters .

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That's the way I like to brain it out. If you prefer to use the formula,
the first problem you run into is: You need to remember the formula !

The formula is D = 1/2 a T²

   Distance = (1/2 acceleration) x (time in seconds)²

   Acceleration = (change in speed) / (time for the change)
    = (18 m/s) / (12 sec)
    = 1.5 m/s² .

Distance = (1/2 x 1.5 m/s²) x (12 sec)²
      =     (0.75 m/s²) x (144 sec²) = 108 meters .

5 0

3 years ago

What terms are needed to completely describe velocity?

kotegsom [21]

In order to completely describe a velocity,
you need a speed and a direction.

4 0

3 years ago

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

8 0

3 years ago

On January 22, 1943, the temperature in Spearfish, South Dakota, rose from -4.0∘F∘F to 45.0∘F∘F in just 2 minutes. What was the

Margaret [11]

Answer:

The change in temperature, $\Delta T=9.45^{\circ} C$

Explanation:

Given that,

The temperature in Spearfish, South Dakota, rose from $-4^{\circ} F\ to\ 45^{\circ} F$ in just 2 minutes. We need to find the temperature change in Celsius degrees. Change in temperature is given by final temperature minus initial temperature such that,

$\Delta T=T_f-T_i\\\\\Delta T=45-(-4)\\\\\Delta T=49^{\circ}F$

The relation between degrees Celsius and degrees Fahrenheit is given by :

$F=1.8C+32$

Here, F = 49 degrees

$49=1.8C+32\\\\\Delta T=9.45^{\circ} C$

So, the change in temperature is 9.45 degree Celsius. Hence, this is the required solution.

8 0

2 years ago

What is the difference between fertilization and pollination

lana [24]

Pollination is when the male gamete or pollen reaches the female flower through different modes...wind,air,animals,...
Fertilization is when the male gamete "join" with the female gamete to become an embryo.
I hope I helped.:)

4 0

3 years ago