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3 years ago

A force of 600 N is acting on a motorcycle that has a mass of 240 kg. What is the acceleration of the motorcycle?

Physics

2 answers:

Verdich [7]3 years ago

7 0

Answer:

2.5m/s2

Explanation:

The following were obtained from the question:

F = 600N

M = 240 kg

a =?

Recall: F = Ma

a = F/M

a = 600/240

a = 2.5m/s2

Therefore, the acceleration of the motorcycle is 2.5m/s2

MaRussiya [10]3 years ago

6 0

Answer:

Explanation:

force(f)=600N

Mass(m)=240kg

Acceleration(a)=?

Acceleration=force/mass

Acceleration=600/240

Acceleration=2.5m/s^2

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What is a synonym & an antonym for property ?<br> What is a synonym of physical change ?

tigry1 [53]

Answer:

A synonym for property is a quality or a characteristic. An antonym is non-physical. A synonym of physical change is a change in state or activity. Phase transition or Natural action.

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2 years ago

Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second

shepuryov [24]

Answer:

distance between both oasis ( 1 and 2) is 27.83 Km

Explanation:

let d is the distance between oasis1 and oasis 2

from figure

OC = 25cos 30

OE = 25sin30

OE = CD

Therefore BC = 30-25sin30

distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem

in $\Delta BCO$

$OB^2 = BC^2 + OC^2$

PUTTING ALL VALUE IN ABOVE EQUATION

$d^2 = 930-25sin30)^2 + (25cos30)^2$

$d^2 = 775$

d = 27.83 Km

distance between both oasis ( 1 and 2) is 27.83 Km

3 0

3 years ago

a 1210 kg roller coaster car is moving 6.33 m/s. as it approaches the station, brakes slow it down to 2.38 m/s over a distance o

Stels [109]

Answer:

4960 N

Explanation:

First, find the acceleration.

Given:

v₀ = 6.33 m/s

v = 2.38 m/s

Δx = 4.20 m

Find: a

v² = v₀² + 2aΔx

(2.38 m/s)² = (6.33 m/s)² + 2a (4.20 m)

a = -4.10 m/s²

Next, find the force.

F = ma

F = (1210 kg) (-4.10 m/s²)

F = -4960 N

The magnitude of the force is 4960 N.

4 0

3 years ago

Read 2 more answers

Calculate the potential difference across a 10 ohm register carrying a current of 2.5 ampere

Bas_tet [7]

Answer:

using ohm's law

V=IR

V= 10 X 0.2

V = 2 Volt.

Explanation:

4 0

2 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago