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3 years ago

9

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

r (mph). At full power, the car can accelerate from zero to 29.0 mph in time 1.10 s .A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0 mph in time 1.10 s , how long would it take to go from zero to 58.0 mph ?

1 answer:

bekas [8.4K]3 years ago

6 0

Answer:

2.2 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Converting mph to m/s

$29\ mph=29\times 0.44704=12.96\ m/s$

$58\ mph=58\times 0.44704=25.93\ m/s$

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{12.96-0}{1.1}\\\Rightarrow a=11.78\ m/s^2$

Considering this acceleration to be constant

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{25.93-0}{11.78}\\\Rightarrow t=2.20\ s$

Time it would take to go from zero to 58.0 mph is 2.2 seconds

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Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

$v_0-gt = 0$ .

where $v_0$ is the initial velocity.

Solving for $t$ we get

$t = \dfrac{v_0}{g}$

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time $t_0$ since it had reached the highest point, the ball has traveled downwards and the velocity $v_f$ it has gained is

$v_f = gt_0$ ,

and we are told that this is twice the initial velocity $v_0$ ; therefore,

$v_f = 2v_0 = gt_0$

which gives

$t_0 = \dfrac{2v_0}{g}.$

Thus, the total time taken to reach velocity $2v_0$ is

$t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}$

$t_{tot} = \dfrac{3v_0}{g}.$

This $t_{tot}$ , we are told, is 36 seconds; therefore,

$36= \dfrac{3v_0}{g},$

and solving for $v_0$ we get:

$v_0 = \dfrac{36g}{3}$

$v_0 = \dfrac{36s(10m/s^2)}{3}$

$\boxed{v_0 = 120m/s}$

which from the options given is choice e.

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Short time large force

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What subatomic particle acts like a mini-magnet?

Semenov [28]

The subatomic particles that acts like a mini-magnet is electron. Electrons are negatively charged sub atomic particles in an atom. The electron spin is a property of an electron that makes it behave like it's spinning; a spinning electron produces a magnetic field that makes it behave like a tiny magnet in an atom.

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A diver makes 1.0 revolutions on the way from a 9.2-m-high platform to the water. Assuming zero initial vertical velocity, find

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Answer:

Average angular velocity ≈ 4.59 rad/s

Explanation:

Using the equation of motion,

H = ut + (1/2)t² ............................ equation 1.

Where H= height, u = initial velocity(m/s), g = acceleration due to gravity(m/s²), t = time(s) u= 0 ∴ ut =0

H =(1/2)gt².................................... equation 2.

making t² the subject of the relation in equation 2,

∴ t² = 2H/g

Where H = 9.2 m, g= 9.8 m/s

∴ t² = ( 2×9.2)/9.8

t = √(2 × 9.2/9.8) = √(18.4/9.8)

t = 1.37 s.

The average angular velocity = θ/t

Where θ = is the number of revolution that the diver makes, t = time

θ = 1 rev.

Since 1 rev = 2π (rad)

t = 1.37 s

Average angular velocity = 2π/t

π = 3.143

Average angular velocity = (2×3.143)/1.37 = 6.286/1.37

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a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

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Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, $M=75\ kg$

Mass of the ball is, $m=4\ kg$

Initial velocity of the man is, $u_m=0\ m/s(rest)$

Initial velocity of the ball is, $u_b=0\ m/s(rest)$

Final velocity of the ball is, $v_b=3.50\ m/s$

Final velocity of the man is, $v_m=?\ m/s$

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of man and ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Now, initial momentum = final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

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3 years ago