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2 years ago

A person is walking down the street maintain the same speed and direction. The person is accelerating.

Physics

1 answer:

balandron [24]2 years ago

6 0

Answer:

no change in speed, therefore the body cannot be accelerated. a=0

Explanation:

When a person is accelerating his speed must change, if the speed is in the same direction as the acceleration the speed increases and if the acceleration is in the opposite direction to the speed it decreases.

In this case there is no change in speed, therefore the body cannot be accelerated.

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if the coefficient of linear expansion of a metal is 2.05× 10^-6 k^-1 what will be its new length if 50cm metal went through a t

boyakko [2]

Answer:

L = L0 (1 + c T) where c is the coefficient and T the change in temperature

L = 50 ( 1 + 2.05E-6 * 50) = 50.0051 cm

7 0

2 years ago

A car that brakes suddenly comes to a screeching halt. Is the sound energy produced in this conversion a useful form of energy?

Nezavi [6.7K]

Answer:

No, the sound produced by screeching brakes is not a useful form of energy, because it does not contribute to stopping the car.

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2 years ago

A 1.2 x 103 kg racecar, with a velocity of 8 m/s, collides with an unsuspecting 80 kg honey badger who is standing

aalyn [17]

Answer: 90 m/s

Explanation:

Given

mass of racecar $M=1.2\times10^3\ kg$

velocity of racecar $u=8\ m/s$

mass of still honeybadger $m=80\ kg$

after collision race car is traveling at a speed of $v_1=2\ m/s$

conserving linear momentum

$Mu+m\times0=Mv_1+ mv_2\quad[v_2=\text{velocity of honeybadger after colllision}]$

$1.2\times10^3\times8+0=1.2\times10^3\times2+80\times v_2$

$1.2\times10^3(8-2)=80v_2\\v_2=\frac{7.2\times10^3}{80}=90\ m/s$

4 0

3 years ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

7 0

3 years ago

Explain the difference between impulse and impulsive force?

ExtremeBDS [4]

An impulsive force is a force that is acting only during a short time, I mean, for an instant. Impulse is a physics magnitude define by the product of the impulsive force and the time that it was acting.

Is there any mistake in my English? Please, let me know.

7 0

2 years ago