All categories

4 years ago

Car accelerates 86km/hr in 4.6s . What is the average acceleration in m/s^2

Physics

1 answer:

never [62]4 years ago

4 0

speed = 86km/hr

= 86000/3600

= 23.889m/s

time = 4.6s

acceleration = 23.889/4.6

a = 5.19m/s^2

You might be interested in

A star having a luminosity 10,000 times that of the sun and a surface temperature only two-thirds of that of the sun, or 4,000k,

ELEN [110]

The hertzsprung-russel diagram is a diagram that correlates the different factors such as effective temperature in Kelvin,absolute mangnitude, relative luminosity of the star compared to the sun and the spectral class of the star. The temperature of the star moves from the left to right that is from blue to red star. To determine the group of stars, we just have to trace the factors in the diagram. We can tell from the diagram that the area surrounding the point are supergiants including Betelguese and Antares. The spectral class is between KO and MO. The class where the star belongs is the supergiants.

7 0

3 years ago

Please guys help me i will help u too

matrenka [14]

Answer:

calculate the cars acceleration usingv=u+at

Explanation:

m/s. After 5 s the car reaches the bottome of the hill. Its speed at the bottom of the ... accelerating left a rownie. 10. A cart slows down while moving away from the ... does it need to accelerate to a velocity of 20 m/s

3 0

3 years ago

Read 2 more answers

Please help I don’t get this give me answers please

Lelu [443]

Answer:

Explanation:

3 0

3 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

4 years ago

The long handle on a rake is a _____.

artcher [175]

From what i know it is c. it is a lever

6 0

3 years ago