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-Dominant- [34]

2 years ago

12

A motorboat starting from rest travels in a straight line on a lake. If the boat reahes a speed of 8.0m/s in 10s what is the boa

t's average acceleration?

1 answer:

victus00 [196]2 years ago

5 0

The boat's average acceleration is 0.8 m/s².

<h3>What is acceleration?</h3>

Acceleration is defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

A motorboat starting from rest travels in a straight line on a lake. If the boat reaches a speed of 8.0m/s in 10s.

The final velocity v = 8 m/s and it starts from rest, then its initial velocity v = 0 m/s.

Time taken for the change in speed t = 10 s.

The acceleration is given by

a = (8 - 0)/ 10

a = 0.8 m/s²

Thus, the boat's average acceleration is 0.8 m/s².

Learn more about acceleration.

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A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.07 Hz. On this tray is an empty cu

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Answer:

Explanation:

Given

Frequency of SHM is $f=2.07\ Hz$

Amplitude of SHM is $A=3.13\ cm$

Cup begins to slip when it overcomes the friction force

Friction force $F_s=\mu mg$

Applied force $F=ma$

$ma=\mu mg$

$a=\mu g$

and maximum acceleration during SHM is

$a=A\omega ^2$

$a=A(2\pi f)^2$

$a=3.13\times 10^{-2}\times (2\pi 2.07)^2$

$a=5.296\ m/s^2$

$\mu =\frac{a}{g}$

$\mu =\frac{5.296}{9.8}=0.54$

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4 years ago

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Why is the process of photosynthesis important to food webs?

Slav-nsk [51]

Answer:

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3 years ago

What are the principals of electric arc welding

Wewaii [24]

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7 0

4 years ago

A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exert

strojnjashka [21]

Answer:

$v=10m/sec$

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

$F_N+Fg$

where

$F_N =forceof the normal$

$Fg= force due to gravity$

Generally the net force is given to be $FC(force towards center)$

$F_C =F_N + Fg$

$F_N =F_C -Fg$

$F_N=F_C-F_g$

$F_N=\frac{mv^2}{R} -mg$

Mathematical we can now derive V

$m_g + \frac{8m}{4}= \frac{mv^2}{8}$

$\frac{5mg}{4} =\frac{mv^2}{8}$

$v^2 =\frac{40*10}{4}$

$v=10m/sec$

Therefore the speed of the roller coaster is given ton be $v=10m/sec$

5 0

3 years ago

What is the gravitational potential energy of a 0.550-kg projectile flying with 335 m/s, 72 meters above the ground?

Fofino [41]

Answer:

GPE = 388.08 Joules.

Explanation:

Given the following data;

Mass = 0.550kg

Speed = 335 m/s

Height = 72 meters

We know that acceleration due to gravity, g is equal to 9.8 m/s²

To find the gravitational potential energy;

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

$G.P.E = mgh$

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

$G.P.E = 0.550 * 9.8 * 72$

GPE = 388.08 Joules.

4 0

3 years ago