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3 years ago

How much work must be done on a 20-kg go-cart to increase its speed from 5 m/s to 10 m/s?

2 answers:

Rudiy273 years ago

5 0

The answer is the change in kinetic energy

K = 1/2 • mv^2

Change in K = (1/2 • 20 • 10^2) - (1/2 • 20 • 5^2)

Change in K = 1000 - 250 = 750J

serious [3.7K]3 years ago

5 0

Answer:

750 J APEX Verified

Explanation:

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A net force of 24 N is acting on a 4.0-kg object. Find the acceleration in m/s2.

Inessa [10]

Hi there!

We can use Newton's Second Law:
$\Sigma F = ma$

ΣF = Net force (N)
m = mass (kg)
a = acceleration (m/s²)

We can rearrange the equation to solve for the acceleration.

$a = \frac{\Sigma F}{m}\\\\a = \frac{24}{4} = \boxed{6 \frac{m}{s^2}}$

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2 years ago

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sveticcg [70]

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3 years ago

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The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,

kondaur [170]

Answer:

$a= -2\ m/s^2$

$a=-12.5\ m/s^2$

x=2.17 m

x=8.4 m

Explanation:

Given that

$v=A+Bt^{-1}$

$v=2+0.25t^{-1}$

To find acceleration :

we know that

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=0-0.5t^{-2}$

$a=-0.5t^{-2}$

Acceleration at t= 2 s

$a=-0.5\times 2^{-2}$

$a= -2\ m/s^2$

Acceleration at t= 5 s

$a=-0.5\times 5^{-2}$

$a=-12.5\ m/s^2$

We know that

$v=\dfrac{dx}{dt}$

$dx=\left(2+\dfrac{1}{4t}\right)dt$

Position at t= 2 s:

$\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{2}$

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

$\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{5}$

x=8+0.25 ln5

x=8.4 m

4 0

3 years ago

A 2.3 kg particle-like object moves in a plane with velocity components vx = 40 m/s and vy = 75 m/s as it passes through the poi

Sonbull [250]

Answer:

(a) $\overrightarrow{L}=885.5\widehat{k}$

(b) $\overrightarrow{L}=1046.5\widehat{k}$

Explanation:

mass, m = 2.3 kg

vx = 40 m/s

vy = 75 m/s

(a) Angular momentum is given by

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

Where, p is the linear momentum and r is the position vector about which the angular momentum is calculated.

Here, $\overrightarrow{r}=3\widehat{i}-4\widehat{j}$

$\overrightarrow{p}=m\overrightarrow{v}$

$\overrightarrow{p}=2.3\left ( 40\widehat{i}+75\widehat{j} \right )$

$\overrightarrow{p}= 92\widehat{i}+172.5\widehat{j}$

So, the angular momentum

$\overrightarrow{L}=\left ( 3\widehat{i}-4\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )$

$\overrightarrow{L}=885.5\widehat{k}$

(b) Here, $\overrightarrow{r}=(3+2)\widehat{i}+(-4+2)\widehat{j}$

$\overrightarrow{r}=5\widehat{i}-2\widehat{j}$

$\overrightarrow{L}=\left ( 5\widehat{i}-2\widehat{j} \right )\times\left ( 92\widehat{i}+172.5\widehat{j} \right )$

$\overrightarrow{L}=1046.5\widehat{k}$

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3 years ago

Definition of complex machine

ASHA 777 [7]

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3 years ago

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