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2 years ago

If an object moving at 5 m/s accelerates for 30 seconds at a rate of 2

Physics

1 answer:

klasskru [66]2 years ago

4 0

Answer:

V=u + at

V= final velocity

U= Initial velocity

a= acceleration

t= time taken

V= 5 + 30*2

V=5+60

V=65m/s

Explanation:

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How much energy is needed to melt 150 g of ice at 0°C to water? (1)(Lf =3.34˟ 10⁵ J/Kg)

NARA [144]

Answer:

5.01×10⁴ J.

Explanation:

Applying,

q = Cm....................... Equation 1

Where q = amount of heat needed to melt the ice, m = mass of the ice, C = specific latent heat of ice.

From the question,

Given: m = 150 g = (150/1000) kg = 0.15 kg, C = 3.34×10⁵ J/kg

Substitute these values into equation 1

q = (0.15×3.34×10⁵)

q = 0.501×10⁵ J

q = 5.01×10⁴ J.

5 0

2 years ago

Based on the chemical equation, use the drop-down menu to choose the coefficients that will balance the chemical

Vadim26 [7]

Answer:

2 1 1

Explanation:

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2 years ago

a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik

grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height of the ball when it hits the ground

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball

$t$ is the time when the ball strikes the ground

$g=9.8m/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $t$ from (1):

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$ (5)

Solving (5) we find the positive result is:

$t=17.68 s$

7 0

2 years ago

Unit 12 Exam

lapo4ka [179]

Answer:

You increase the acceleration of the car.

6 0

2 years ago

Martha must carry a 45 N package up three flights of stairs. Each flight of stairs has a height of 2.3m, and the actual distance

Darina [25.2K]

Answer:

310.5 J

Explanation:

The total work done by Martha is equal to the increase in gravitational potential energy of the package, which is equal to

$\Delta U = mg\Delta h$

where

(mg) = 45 N is the weight of the package

$\Delta h$ is the increase in height of the package

The package is carried up 3 flights of stairs, each one with a height of 2.3 m, so the total increase in heigth is

$\Delta h = 3 \cdot 2.3 m=6.9 m$

And so, the work done by Martha is

$U=(45 N)(6.9 m)=310.5 J$

3 0

3 years ago