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3 years ago

14

A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does

the boat travel during this time

1 answer:

ankoles [38]3 years ago

4 0

Answer:

96m

Explanation:

Using SUVAT:

s=? u=0 a=3 t=8

s=ut+0.5*at^2

s=0.5*3*8^2

s=96m

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The end diastolic volume of a heart is 140 mL Assume that it is a sphere. At end diastole, the intraventricular pressure is 7mmI

Vera_Pavlovna [14]

Answer:

Explanation:

We know that, V = 140 mL = 0.00014 m3

Assume that it is a sphere. so, we have

V = (4/3) \pir3

r3 = (0.00014 m3) (3) / (4) (3.14)

r = \sqrt[3]{}\sqrt[3]{}3\sqrt{}3.34 x 10-5 m3

r = 1.93 x 10-7 m

(a) The wall tension at end diastole will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end diastole = 7 mmHg = 933.2 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (933.2 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 8.18 x 10-3 N

(b) The wall tension at the end of isovolumetric contraction will be given as :

using a formula, we have

T = P r / 2 H

where, P = intraventricular pressure at end of isovolumetric contraction = 80 mmHg = 10665.7 Pa

H = wall thickness at this time = 0.011 m

then, we get

T = (10665.7 Pa) (1.93 x 10-7 m) / 2 (0.011 m)

T = 9.35 x 10-2 N

(d) The wall stress from A and B which will be given as :

we know that, \sigma = T / w

For part A, we have

\sigmaA = (8.18 x 10-3 N) / (0.011 m)

\sigmaA = 0.743 N/m

For part B, we have

\sigmaB = (9.35 x 10-2 N) / (0.011 m)

\sigmaB = 8.5 N/m

4 0

3 years ago

In SI units, what is the magnitude the net force acting on a 1,152 kg car that accelerates uniformly along a straight line from

geniusboy [140]

Answer:

Fnet = 14515.2 Newton

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

$Fnet = Fapp + Fg$

Where;

Fnet is the net force.
Fapp is the applied force.
Fg is the force due to gravitation.

Given the following data;

Mass = 1,152 kg

Initial velocity, u = 3m/s

Final velocity, v = 17m/s

Time, t = 5 seconds

To find the magnitude of the net force;

First of all, we would determine the acceleration of the car.

Acceleration = (v-u)/t

Acceleration = (17 - 3)/5

Acceleration = 14/5

Acceleration = 2.8m/s

To find the applied force;

Fapp = mass * acceleration

Fapp = 1,152 * 2.8

Fapp = 3225.6 N

Next, we would find the force exerted on the car due to gravity.

Fg = mass * acceleration due to gravity

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

Fg = 1152 * 9.8

Fg = 11289.6N

Substituting the values into the equation, we have;

Fnet = 3225.6 + 11289.6

Fnet = 14515.2 Newton

7 0

3 years ago

How many neutrons are needed to initiate the fission reaction shown?

nalin [4]

Answer: One neutron

Explanation:

one neutron 1/0n

Sum up the mass numbers on the right 99 + 135 + 2 = 236.

The sum of the mass numbers on the left should equal 236. 235 + 1 = 236

4 0

3 years ago

Is the Force balanced or unbalanced?

Alexus [3.1K]

Answer:

1. balanced

2. unbalanced

3. balanced

4. balanced

5. unbalanced

6. unbalanced

7. unbalanced

8. balanced

9. unbalanced

10. unbalanced

11. unbalanced

12. balanced

3 0

3 years ago

A biker pedals hard to ride his bike to the top of a 44 m hill. He decides to let his bike coast down the hill, and is having so

Dafna11 [192]

Answer:

The bikers speed at the top of other hill is <u>25.82 m/s.</u>

Explanation:

Considering the biker is riding on a frictionless surface.

∴ There is no non-conservative or external force acting on the biker.

Hence we can conserve the energy of biker and bike as a system.

Let,

$h_{1}$ = 44m

$h_{2}$ = 10m

Since the biker starts from rest , his initial speed $v_{1}$ = 0 m/s

Let final speed of the bike at the top of other hill be $v_{2}$ .

∴ Initial Energy (at the top of 44m hill) = $mgh_{1}$

Final Energy (at the top of 10m hill) = $mgh_{2} + \frac{1}{2}mv_{2} ^{2}$ .

Conserving both the energies , we get

$mgh_{1}$ = $mgh_{2} + \frac{1}{2}mv_{2} ^{2}$

∴ $v_{2} = \sqrt{2g(h_{1}-h_{2} )}$

Substituting the values for g , $h_{1}$ , $h_{2}$ , we get

$v_{2}$ = 25.82 m/s

6 0

3 years ago