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3 years ago

A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?

Physics

2 answers:

OLga [1]3 years ago

7 0

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

marusya05 [52]3 years ago

6 0

Answer : The correct option is, (A) 7.5 mN.s

Explanation :

Impulse : It is defined as the force applied to an object for a certain amount of time.

Formula used :

$I=F\times \Delta t$

As we know that:

$F=m\times \Delta a\\\\\Delta a=\frac{\Delta v}{\Delta t}$

Thus, formula of impulse will be:

$I=m\times \Delta v$

where,

I = impulse = ?

m = mass = 0.15 kg

$\Delta v$ = change in speed of velocity = $v_{final}-v_{initial}=15cm/s-10cm/s=5cm/s=0.05m/s$ (1 m = 100 cm)

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

$I=(0.15kg)\times (0.05m/s)$

$I=0.0075kg.m/s=0.0075N.s=7.5mN.s$ (1 N.s = 1000 mN.s)

Therefore, the impulse acting on the car is, 7.5 mN.s

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2 years ago

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lana66690 [7]

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3 years ago

8. The ratio of output power to input power, in percent?A. work. B. efficiency. C. horsepower. D. conductivity.

Tema [17]

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3 years ago

Read 2 more answers

Timmy bikes to school 3 kilometers east and 2 kilometers south. What is Timmy's final displacement?

vovikov84 [41]

Answer:

3.6km South East

Explanation:

Displacement is the shortest distance between the starting point and the ending point and the direction it is displaced in. To calculate the displacement we can use the pythagoras theorem because the 3km East and the 2km south form the two shorter sides of a right angled triangle between the starting and ending points. So, the displacement is the length C of the triangle which we can calculate as follows:

Pythagoras Theorem:

a^2+b^2=c^2

(2)^2+(3)^2=c^(2)

4+9=c^2

Square root 13 = c

c=3.6km (1dp)

The total displacement is 3.6km and is in the approximate direction of South East (because he travelled east and south).

Hope this helped!

8 0

2 years ago

Kūna veikia jėga kurios momentas 0,4 N × m petys - 5 cm. Koks šios jėgos dydis?

umka21 [38]

Answer:

F = 8 N

Explanation:

The question says "The body is subjected to a force with a moment of 0.4 N × m shoulder - 5 cm. What is the magnitude of this force?

Given that,

Moment/Torque, $\tau=0.4\ N-m$

Distance moved, d = 5 cm = 0.05 m

We need to find the magnitude of this force. We know that, the torque acting on an object is given by :

$\tau=Fd\\\\F=\dfrac{\tau}{d}\\\\F=\dfrac{0.4\ N-m}{0.05\ m}\\\\F=8\ N$

So, the magnitude of force is equal to 8 N.

8 0

2 years ago