A force of 240.0 N causes an object to accelerate at 3.2 m/s2. What is the mass of the object?

Answers

Answer 1

Answer: the mass would be 75kg

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Two particles, one with charge -6.29 × 10^-6 C and one with charge 5.23 × 10^-6 C, are 0.0359 meters apart. What is the magnitude of the force that one particle exerts on the other?
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A basketball player jumps straight up for a ball. To do this, he lowers his body 0.250 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.960 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor. 4.33 Correct: Your answer is correct. m/s (b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.

A person exerts a horizontal force of F=45N on the end of an 86cm wide door. The magnitude of the torque due to F about the pivot point is determined by |τ|=|rxF|=rFsinθ . Determine the magnitude of the torque, |τ| , on the door about its hinges due to F . |τ|=0Nm |τ|=38.7Nm |τ|=3870Nm

Answers

Answer: The magnitude of torque is 38.7Nm

Explanation: Please see the attachment below

The magnitude of the torque on the door about its h1nges due to the applied force is 38.7 Nm.

How to calculate the magnitude of the torque?

The magnitude of the torque on the door about its h1nges due to the applied force is calculated by applying the following formula as shown below;

τ = rF

where;

r is the perpendicular distance of the applied force
F is the applied force

The given parameters include;

perpendicular distance, r = 86 cm = 0.86 m

the applied force , F = 45 N

The magnitude of the torque on the door about its h1nges due to the applied force is calculated as;

τ = rF

τ = 0.86 m x 45 N

τ = 38.7 Nm

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n a distant solar system, a planet of mass 5.0 x 1024 kg orbits a sun of mass 3.0 x 1030 kg at a constant distance of 2.0 x1011 m. How many earth days does it take for the planet ot execute one complete orbit about the sun

Answers

Answer:

F = M2 ω^2 R centripetal force of sun on planet

ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P where P is the period

P = 2 pi (M2 * R / F)^1/2

F = G M1 M2 / R^2 gravitational force on planet

P = 2 pi {R^3 / (G M1)]^1/2

P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2

P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec

1 yr = 3600 * 24 * 365 = 3.15E7 sec

P = 3.9 / 3.2 = 1.2 years

Make the following conversion.56.32 kL = _____ L

0.056320
0.56320
5,632
56,320

Answers

Answer:

The answer would be D 56,320

Explanation:

A block-spring system consists of a spring with constant k = 475 N/m attached to a 2.50 kg block on a frictionless surface. The block is pulled 5.50 cm from equilibrium and released from rest. For the resulting oscillation, find the amplitude, angular frequency, frequency, and period. What is the maximum value of the block's velocity and acceleration?

Answers

Explanation:

It is given that,

Spring constant of the spring, k = 475 N/m

Mass of the block, m = 2.5 kg

Elongation in the spring from equilibrium, x = 5.5 cm

(a) We know that the maximum elongation in the spring is called its amplitude. So, the amplitude for the resulting oscillation is 5.5 cm.

(b) Let $\omega$ is the angular frequency. It is given by :

$\omega=\sqrt{(k)/(m)}$

$\omega=\sqrt{(475)/(2.5)}$

$\omega=13.78\ rad/s$

(c) Let T is the period. It is given by :

$T=(2\pi)/(\omega)$

$T=(2\pi)/(13.78)$

T = 0.45 s

(d) Frequency,

$f=(1)/(T)$

$f=(1)/(0.45)$

f = 2.23 Hz

(e) Let v is the maximum value of the block's velocity. It is given by :

$v_(max)=\omega* A$

$v_(max)=13.78* 5.5* 10^(-2)$

$v_(max)=0.757\ m/s$

The value of acceleration is given by :

$a=\omega^2A$

$a=(13.78)^2* 5.5* 10^(-2)$

$a=10.44\ m/s^2$

Hence, this is the required solution.

Final answer:

The amplitude of the block-spring system is 0.055 m, with an angular frequency of 13.77 rad/s, and a frequency of approximately 2.19 Hz. The system has a period of approximately 0.46 s, with the maximum velocity being 0.76 m/s, and the maximum acceleration being 189 m/s².

Explanation:

In this block-spring system, we can determine the oscillation properties with the given parameters: mass (m = 2.50 kg), spring constant (k = 475 N/m), and displacement (x = 5.50 cm = 0.055 m).

Amplitude (A): This is the maximum displacement from the equilibrium position, in this case, it is 0.055 m, the distance from which the block is pulled and released.
Angular Frequency (ω): This can be calculated by the formula ω = √(k/m) which is equal to √(475/2.50) = 13.77 rad/s.
Frequency (f): It is given by the formula f = ω / 2π ≈ 2.19 Hz.
Period (T): The time for one complete cycle of the motion, calculated as T = 1/f ≈ 0.46 s.
Maximum value of velocity (v max): Calculated by the formula v = ω*A = 13.77 * 0.055 = 0.76 m/s.
Maximum value of acceleration (a max): Maximum acceleration occurs when the block is at the maximum displacement (amplitude), found with formula a = ω²*A = 189 m/s².

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An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Determine the speed at this time and the maximum height at which it reaches.

Answers

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2* 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+(at^2)/(2)$

$s=0+(2* 6^2)/(2)$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=(144)/(2* 9.8)=7.34 m$

$s+s_0=36+7.34=43.34 m$

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.

Answers

Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

$Speed = (Distance)/(Time)$

$Speed = (5)/(0.504)$

= 9.92 m/s

Now

$v^2 - u^2 = 2* g* h$

$9.92^2 = 2* 9.98 * h$

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Final answer:

The height of the water slide is 5.04 meters.

Explanation:

The problem described in this question involves a water slide, where swimmers start from rest at the top and leave the slide traveling horizontally. To determine the height of the slide, we can use the equations of motion in the horizontal direction. The horizontal displacement (x) is given as 5.00 m and the time (t) is given as 0.504 s. Assuming no friction or air resistance, we can use the equation x = v*t, where v is the horizontal velocity. Rearranging the equation, we can solve for v, which is equal to x/t. Substituting the given values, we have v = 5.00 m / 0.504 s = 9.92 m/s. The horizontal velocity (v) is constant throughout the motion, so we can use the equation v = sqrt(2*g*H), where g is the acceleration due to gravity (9.8 m/s^2) and H is the height of the slide. Rearranging the equation, we can solve for H, which is equal to v^2 / (2*g). Substituting the known values, we have H = (9.92 m/s)^2 / (2*9.8 m/s^2) = 5.04 m.

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