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3 years ago

Saeed moves 50 m east then 65 m south to reach the garden. What is the displacement of Saeed?

Physics

1 answer:

vampirchik [111]3 years ago

4 0

Answer:

82.01m

Explanation:

Using the formula:

R^2=A^2+B^2

R= √ A^2+B^2

Where R= displacement

A=50m

B=65m

R= √ 50^2+65^2

R= √ 2500+4225

R= √ 6725

R=82.01m

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According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this

Andrew [12]

Answer:

Part a)

$\rho = 0.55 g/cm^3$

Part b)

$\rho_L = 1.49 g/cm^3$

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

Explanation:

Part a)

As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force

So here we know

buoyancy force is given as

$F_b = \rho_w V_{sub} g$

$F_b = (1 g/cm^3) (30 - 13.5) Ag$

$F_b = 16.5 Ag$

Now we know that the weight of the cylinder is given as

$W = \rho (30 cm)A g$

now we have

$\rho (30 cm) A g = 16.5 A g$

$\rho = 0.55 g/cm^3$

Part b)

When the same cylinder is floating in other liquid then we will have

$F_b = \rho_L (30 - 18.9 )A g$

so we have

$\rho_L (11.1) Ag = 0.55(30) Ag$

$\rho_L = 1.49 g/cm^3$

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

3 0

3 years ago

Starting from rest, a 4.0-kg body reaches a speed of 8.0 m/a in 20 s. What is te net force acting on the body?

Strike441 [17]

The guy below is wrong!

F=ma
Where force = mass x acceleration

We dont have acceleration, a= change in velocity divided by the time taken.
a = v (final velocity) - u (initial) / t
a us 8-0 (at rest means u was 0) / 20 = 0.4

Using F=ma

F= mass x acceleration
F= 4 x 0.4
F=1.6 N

5 0

3 years ago

Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts

podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height =?

$y_{0}$ : is the initial height = 1 m

$v_{0_{y}$ : is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time

First, we need to find the time by using the following equation:

$t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s$

Now, the height is:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m$

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

$0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}$

$1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0$

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

$x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m$

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!

4 0

3 years ago

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J = 120 J decrease

The missing energy has turned into interned energy in the completely inelastic collision

3 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

4 years ago

Saeed moves 50 m east then 65 m south to reach the garden. What is the displacement of Saeed?​

Saeed moves 50 m east then 65 m south to reach the garden. What is the displacement of Saeed?