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2 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]2 years ago

7 0

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

p=mv=0.15•5=0.75 kg•cm/s

marusya05 [52]2 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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DENIUS [597]

Answer:

$v_{oy}=11.29\ m/s$

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

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2 years ago

You are pulling your little sister on her sled across an icy (frictionless) surface. When you exert a constant horizontal force

Tpy6a [65]

Answer:

Mass of Little Sister = 44.17 kg

Explanation:

From Newton's second law of motion, the magnitude of force applied on the sled is given by the following formula:

F = ma

where,

F = Force Applied = 120 N

a = Acceleration = 2.3 m/s²

m = Mass of Sled + Mass of Little Sister = 8 kg + Mass of Little Sister

Therefore,

120 N = (2.3 m/s²)(8 kg + Mass of Little Sister)

(120 N)/(2.3 m/s²) = 8 kg + Mass of Little Sister

Mass of Little Sister = 52.17 kg - 8 kg

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2 years ago

A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod

ivann1987 [24]

Answer:

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Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

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Now,

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$e = - l(vec{v}\times \vec{B})$

Here, velocity v is perpendicular to the rod

Thus

e = lvB (1)

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$v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}$

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Divide each side by (voltage): Power / voltage = Current

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2 years ago