All categories

3 years ago

How do you determine the acceleration of an object?

Physics

2 answers:

Mekhanik [1.2K]3 years ago

7 0

according to the second law of dynamics F = m • a => a = F / m

Anton [14]3 years ago

5 0

(Final - initial Speed) / time

Hope this helped★

You might be interested in

A cube of mass m1=9.6 kgm1=9.6 kg is sitting on top of a second cube of the same size and mass m2=0.8 kgm2=0.8 kg while both are

GalinKa [24]

Answer:N=0

Explanation:

Given

$m_1=9.6 kg$

$m_2=0.8 kg$

both blocks experiencing free fall so net weight of block during free fall is zero thus there is no normal reaction between them.

N=0

8 0

3 years ago

The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2

sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed $V_{0}$

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}$

Once we have the initial speed, we can isolate the final speed from following equation:

$V_{f} =V_{0} +a*t$ $V_{f}= 4.165 \frac{m}{s}$

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

$x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }$

3 0

3 years ago

Read 2 more answers

What is the next step if the data from an investigation do not support the original hypothesis? A. The data are revised to suppo

Leno4ka [110]

I believe the answer is D. <span>The hypothesis is revised and another experiment is conducted.</span>

5 0

3 years ago

Read 2 more answers

6000 kg train moving 5 m/sec to east collides with 5000 kg train moving 3 m/sec to west. What is their velocity

Svetlanka [38]

The final velocity is 1.37 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 6000 kg$ is the mass of the first train

$u_1 = 5 m/s$ is the initial velocity of the first train (we take east as positive direction)

$m_2 = 5000 kg$ is the mass of the second train

$u_2 = -3 m/s$ is the initial velocity of the second train

$v$ is the final combined velocity of the two trains

Re-arranging the equation and substituting the values, we find:

$v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(6000)(5)+(5000)(-3)}{6000+5000}=1.37 m/s$

And the positive sign indicates their final direction is east.

Learn more about momentum here:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

8 0

3 years ago

A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple s

posledela

Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

$P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m$

The spring constant of the net is 20130.76 N

From Hooke's Law

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m$

The net would strech 0.03167 m

If h = 35 m

From energy conservation

$65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0$

Solving the above equation we get

$x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45$

The compression of the net is 1.52 m

4 0

3 years ago