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Yanka [14]
3 years ago
5

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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An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F
ololo11 [35]

Answer

given,

mass of jogger  = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed  = 1.3 m/s  in  61 ° north of east.

jogger one

P_1 = m_1 v_1 \hat{i}

P_1 = 67 \times 2.3\hat{i}

P_1 = 154.1 \hat{i}

P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}

P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}

P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}

P_2 = 44.12\hat{i} +79.59\hat{j}

now

P = P₁ + P₂

P = 198.22 \hat{i} +79.59 \hat{j}

magnitude

P = \sqrt{198.22^2 + 79.59^2}

P =213.60 kg.m/s

\theta = tan^{-1}\dfrac{79.59}{198.22}

\theta = 21.87

the angle is \theta = 21.87 north of east

7 0
3 years ago
3<br><br><br> help with physics!!<br><br> ----------------<br><br> will give brainliest
Zinaida [17]

Answer:

Answer B

Explanation:

An increase in resistance makes it harder for the electric current to pass through

6 0
2 years ago
the imagine above shows to opposite forces acting on a rolling cart, what can we say is true about affect of the forces on the c
cluponka [151]

Answer:

Here is my answer...

Explanation:

The cart will connect with the opposite force, and then the cart will come to a shuddering stop before moving in the direction of the oposite force.

Hope I helped! :)

7 0
2 years ago
a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

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

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

4 0
3 years ago
Can y'all please help me out with this ?
Inessa05 [86]

Answer:

no where is the main part of the question dude

3 0
3 years ago
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