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3 years ago

A speedboat initially at rest accelerates uniformly at 4.0 m/s (squared) for 7.0 s. How fast is the boat moving after 7.0 s?

1 answer:

8 0

Well if the boat initially at rest accelerates at uniformly at 4.0 m/s (squared) then it would be best to muitlply it so 4.0 squared equals 2 by multiplying that by 7.0 your answer would be 14 s

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A pulling force is called what? A. normal B. tension C. balanced D. compression

joja [24]

There is no special name for that. Physics is usually just concerned with "forces", and doesn't specify whether the force pushes or pulls. If you want to be more specific, you can just call it a "pulling force".
I hoped this was satisfying!:)

3 0

3 years ago

Read 2 more answers

Over a short interval the coordinate of a car in the meters isgiven by x(t) = 27t - 4.0 t3 where time t is in seconds,at the end

wariber [46]

Answer:

5. -24 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s².

mathematically,

a = dv/dt ............................ Equation 1

Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.

But

v = dx(t)/dt

Where,

x(t) = 27t-4.0t³...................... Equation 2

Therefore, differentiating equation 2 with respect to time.

v = dx(t)/dt = 27-12t²............. Equation 3.

Also differentiating equation 3 with respect to time,

a = dv/dt = -24t

a = -24t .................... Equation 4

from the question,

At the end of 1.0 s,

a = -24(1)

a = -24 m/s².

Thus the acceleration = -24 m/s²

The right option is 5. -24 m/s²

4 0

3 years ago

In tae-kwon-do, a hand is slammed down onto a target at a speed of 10 m/s and comes to a stop during the 7.0 ms collision. Assum

wlad13 [49]

Answer:

(A) Impulse = 9Ns

(B) F = 1286N

Explanation:

Impulse = change in momentum = m(v-u)

v = 0 (the hand comes to a stop)

u = -10m/s

Mass = 0.9kg

Impulse = 0.9 ×(0- (-10))

= 9Ns

(B) F×t = Impulse

F = Impulse/ t

t = 7ms = 7×10-³

F = 9/ (7×10-³)

F = 1286N.

5 0

3 years ago

The animation shows a ball which has been kicked upward at an angle. Run the animation to watch the motion of the ball. Click in

sergejj [24]

Answer:

The acceleration of the ball is $a_y = - 0.3672 \ m/s^2$

Explanation:

From the question we are told that

The maximum height the ball reachs is $H_{max} = 42.24 \ m$

The horizontal component of the initial velocity of the ball is $v_{ix} = 5.57 \ m/s$

The vertical component of the initial velocity of the ball is $v_{iy} = = 16.18 m/s$

The vertically motion of the ball can be mathematically represented as

$v_{fy}^2 = v_{iy} ^2 + 2 a_{y} H_{max}$

Here the final velocity at the maximum height is zero so $v_{fy} = 0 \ m/s$

Making the acceleration $a_y$ the subject we have

$a_y = \frac{v_{iy} ^2}{2H_{max}}$

substituting values

$a_y = - \frac{5.57^2}{2* 42.24}$

$a_y = - 0.3672 \ m/s^2$

The negative sign shows that the direction of the acceleration is in the negative y-axis

6 0

3 years ago

A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti

uysha [10]

Answer:The sled slides 16.875m before rest.

Explanation:

$a=\frac{F}{m} =\frac{12N}{20kg}$

a=0.6 m/s²

$Vf=0=Vi-a.t$

$t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg$

$d= Vi.t - \frac{a.t^{2}}{2}$

$d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m$

3 0

4 years ago