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2 years ago

A child tugs on a rope attached to a 0.62-kg toy with a horizontal force

Physics

1 answer:

Katarina [22]2 years ago

6 0

Answer:

$a=0.8\ m/s^2$

Explanation:

Given that,

Mass of a toy, m = 0.62 kg

Horizontal force, F = 16.3 N

Force on the opposite direction, F' = 15.8 N

We need to find the acceleration of the toy.

Here, two forces are acting in opposite direction, the net force will be the difference of forces.

Net force = 16.3 N-15.8 N

=0.5 N

The formula for net force is given by :

F = ma

a is the acceleration of the toy

$a=\dfrac{F}{m}\\\\a=\dfrac{0.5\ N}{0.62\ kg}\\\\a=0.8\ m/s^2$

So, the acceleration of the toy is $0.8\ m/s^2$ .

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A certain amusement park ride consists of a large rotating cylinder of radius R=3.05 m.R=3.05 m. As the cylinder spins, riders i

aniked [119]

Answer:

a. N = 2.49W b. 0.40

Explanation:

a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?

Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s

Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m

The rider's weight W = mg = 9.8m

The ratio of the normal force to the rider's weight is

N/W = 24.43m/9.8m = 2.49

So the normal force expressed in term's of the rider's weight is

N = 2.49W

b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?

The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.

Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.

So, the normal force equals

N = F/μ = W/μ = mg/μ = mrω²

μ = mg/mrω²

= W/N

= 9.8m/24.43m

= 0.40

6 0

2 years ago

In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d

Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

$x*g = x * \frac{dv}{dt} + v^2$

- Where, v = Speed of the raindrop

- Proposed solution to given ODE:

v = a*t

Where, a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

a = dv / dt ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

v = dx / dt , v = a*t ... proposed solution (Eq 2)

v .dt = dx = a*t . dt

- integrate both sides:

∫a*t . dt = ∫dt

x = 0.5*a*t^2 ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

= 1.5 a^2 t^2

a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

x (t) = 0.5*a*t^2

x (t = 3.0) = 0.5*9.81*3^2 / 3

x (3.0) = 14.7 m

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

m (t) = k*x (t)

m (3.0) = 2.00*x(3.0)

m (3.0) = 2.00*14.7

m (3.0) = 29.4 g

6 0

2 years ago

If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 14m/s, how far from the base of the dorm should

liubo4ka [24]

Emily throws the ball at 30 degree below the horizontal

so here the speed is 14 m/s and hence we will find its horizontal and vertical components

$v_x = 14 cos30 = 12.12 m/s$

$v_y = 14 sin30 = 7 m/s$

vertical distance between them

$\delta y = 4 m$

now we will use kinematics in order to find the time taken by the ball to reach at Allison

$\delat y = v_y *t + \frac{1}{2} at^2$

here acceleration is due to gravity

$a = 9.8 m/s^2$

now we will have

$4 = 7 * t + \frac{1}{2}*9.8 * t^2$

now solving above quadratic equation we have

$t = 0.44 s$

now in order to find the horizontal distance where ball will fall is given as

$d = v_x * t$

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

$d = 12.12 * 0.44 = 5.33 m$

so the distance at which Allison is standing to catch the ball will be 5.33 m

8 0

2 years ago

a falling skydiver slows from a speed of 52 m/s to 8 m/s in 0.8 sec as the parachute opens. what is the diver's acceleration and

Cloud [144]

I found the answers here. Hope this helps you! https://1.cdn.edl.io/sJTle6yxt3qVq7jHfdHRZJ3Xogj7ps6swBO9umNcZ6PO3SMN.docx

8 0

2 years ago

How many top quark lifetimes have there been in the history of the universe (i.e., what is the age of the universe divided by th

ololo11 [35]

Answer:

$1.0\cdot 10^{41}$ times

Explanation:

First of all, we need to write both the age of the universe and the lifetime of the top quark in scientific notation.

Age of the universe:

$T=100,000,000,000,000,000s = 1.0\cdot 10^{17} s$ (1 followed by 17 zeroes)

Lifetime of the top quark:

$\tau = 0.000000000000000000000001s = 1.0\cdot 10^{-24} s$ (we moved the decimal point 24 places to the right)

Therefore, to answer the question, we have to calculate the ratio between the age of the universe and the lifetime of the top quark:

$r = \frac{T}{\tau}=\frac{1.0\cdot 10^{17} s}{1.0\cdot 10^{-24} s}=1.0\cdot 10^{41}$

6 0

2 years ago