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Firdavs [7]
3 years ago
5

The launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m, where

should a target be placed so that the toy car lands on it?
Physics
1 answer:
lora16 [44]3 years ago
7 0

Answer:

1.6 m

Explanation:

Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.

The time for landing should be calculated by using the second equation of motion formula

h = Ut + 1/2gt^2

Let U = 0

0.5 = 1/2 × 9.8 × t^2

0.5 = 4.9t^2

t^2 = 0.5 / 4.9

t^2 = 0.102

t = 0.32 s

The target should be placed so that the toy car lands on it at:

Distance = 5 × 0.32

distance = 1.597 m

Distance = 1.6 m

Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.

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3 years ago
Which of these actions will increase friction? Select three options. Scratching a surface to make it rougher polishing a surface
k0ka [10]

Answer:

scratching a surface to make it rougher

increasing the size of a flying object

adding extra weight to an object

Explanation:

5 0
3 years ago
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Josh did an experiment recording the changes in temperature in sand and water when exposed to a light source, and then when the
Marrrta [24]

Before going to solve this question first we have to understand specific heat capacity of a substance .

The specific heat of a substance is defined as amount of heat required to raise the temperature of 1 gram of substance through one degree Celsius. Let us consider a substance whose mass is m.Let Q amount of heat is given to it as a result of which its temperature is raised  from T to T'.

Hence specific heat  of a substance is calculated as-

                                              c= \frac{Q}{m[T'-T]}

Here c is the specific heat capacity.

The substance whose specific heat capacity is more will take more time to be heated up to a certain temperature as compared to a substance having low specific heat which is to be heated up to the same temperature.

As per the question John is experimenting on sand and water.Between sand and water,water has the specific heat 1 cal/gram per degree centigrade which is larger as compared to sand.Hence sand will be heated faster as compared to water.The substance which is heated faster will also cools faster.

From this experiment John concludes that water has more specific heat as compared to sand.

7 0
3 years ago
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Titanium is a metal used to make golf clubs. A rectangular bar of this metal measuring 1.96 cm x 2.19 cm x 2.63 cm was found to
masha68 [24]

4.012\frac{gr}{cm^3}

Explanation

the density of an object is given by:

\text{Density(d)}=\frac{mass(m)}{\text{volume(v)}}

Step 1

find the volume of the bar

a)find the volume of the rectangular bar.

the volume of a rectangular prism is given by:

\text{Volume}=\text{ length}\cdot widht\cdot depth

replace

\begin{gathered} \text{Volume}=(\text{ 2.63}\cdot2.19\cdot1.96)(cm^3) \\ \text{Volume}=11.289012(cm^3) \end{gathered}

Step 2

now,

Let

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replace in the formula

\begin{gathered} \text{Density(d)}=\frac{mass(m)}{\text{volume(v)}} \\ d=\frac{45.3\text{ gr}}{11.289012(cm^3)} \\ d=4.012\frac{gr}{cm^3} \end{gathered}

therefore, the answer is

4.012\frac{gr}{cm^3}

I hope this helps you

4 0
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navik [9.2K]
It would be static friction which is what you have to overcome when an object is not in motion. When you move an object friction works against it like gravity and air resistance. I hope this helps!
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