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2 years ago

An object accelerates at 32 m/s² when a force of 71 N is applied to it. What is the object’s mass?

Physics

2 answers:

lawyer [7]2 years ago

5 0

Answer:

2.2175 is the answer of this question

GalinKa [24]2 years ago

4 0

Answer:

2.21875 kg

Explanation:

FORCE = MASS x ACCELERATION so to find mass we divide both sides by acceleration. M=F/A

SO 71N/32m/s²=2.21875 kg.

remember N=kgm/s² so m/s² is canceled by m/s² the remaining is kg

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Imagine that the above hoop is a tire. the coefficient of static friction between rubber and concrete is typically at least 0.9.

Stels [109]

The hoop is attached.

Consider that the friction force is given by:
F = μ·N
 = μ·m·g·cosθ

We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·g·sinθ

Therefore:
-(1/2)·m·g·sinθ + m·g·sinθ = μ·m·g·cosθ
(1/2)·m·g·sinθ = μ·m·g·cosθ
μ = (1/2)·m·g·sinθ / m·g·cosθ
 = (1/2)·tanθ

Now, solve for θ:
θ = tan⁻¹(2·μ)
 = tan⁻¹(2·0.9)
 = 61°

Therefore, the maximum angle you could ride down without worrying about skidding is 61°.

5 0

2 years ago

While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe

kaheart [24]

Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

$P_{Miy}$ + $p_{Piy}$ + $j_{y}$ = $P_{Mfy}$ + $P_{pfy}$

$m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sin$ Ф

60 ·8 + 0 = (60 + 80) $V_{f}sin$ Ф

480 = 140 $V_{f} sin$ Ф................. (I)

y-axis component form (+y north);

$P_{Mix}$ + $p_{Pix}$ + $j_{x}$ = $P_{Mfx}$ + $P_{pfx}$

$m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cos$ Ф

0 + 80.9 = (60 + 80) $V_{f}cos$ Ф

720= 140 $V_{f}cos$ Ф

140Vf= $\frac{720}{cos}$ Ф......................................(2)

Substituting (2) into (1) to give the angle;

480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

480 = 140Vfsin 33.69°

Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

7 0

2 years ago

Two 8.0 Ω lightbulbs are connected in a 12 V series circuit. What is the power of both glowing bulbs?

V125BC [204]

Answer:

18 W

Explanation:

Applying,

P = V²/R.................. Equation 1

Where P = Power of both glowing bulbs, V = Voltage, R = Combined Resistance of both bulbs

Since: It is a series circuit,

Then,

R = R1+R2............. Equation 2

Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb

Given: R1 = R2 = 8 Ω

Substitute into equation 1

R = 8+8

R = 16 Ω

Also Given: V = 12 V

Substitute into equation 1

P = 12²/8

P = 144/8

P = 18 W

7 0

2 years ago

A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g an

leonid [27]

Answer:

A) W = 103.55 N

B) mass of displaced water = 4186 g

C) W_displaced water = 41.06 N

D) Buoyant force = 41.06 N.

E) ZERO

F) 62.54 N

Explanation:

We are given;

mass of statuette;m = 10,566 g = 10.566 kg

volume = 4,064 cm³

Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³

A) The dry weight of the statuette can be calculated as;

W = mg

So;

W = 10.556 × 9.81

W = 103.55 N

B) Mass of displaced water is calculated from;

Density = mass/volume

So, mass = Density × Volume

m = 1.03 × 4,064 = 4186 g

C) Weight of displaced water is given by;

W_displaced water = (m_displaced water) × g

W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N

D) The buoyant force is the same as the weight of the displaced water.

Thus, Buoyant force = 41.06 N.

E) The apparent weight of the statuette is calculated from;

Apparent weight = Dry weight - Weight of displaced water

Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.

F. From E above, The Force required to lift the statuette = 62.54 N

4 0

2 years ago

A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 75.0 m high.How much later does it reach t

Katarina [22]

Answer:

5.72 seconds

848.27 m/s

97.94 m

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=15-9.81\times t\\\Rightarrow \frac{-15}{-9.81}=t\\\Rightarrow t=1.52 \s$

Time taken to reach maximum height is 0.97 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=15\times 1.52+\frac{1}{2}\times -9.81\times 1.52^2\\\Rightarrow s=11.47\ m$

So, the stone would travel 11.47 m up

So, total height stone would fall is 75+11.47 = 86.47 m

Total distance travelled by the stone would be 75+11.47+11.47 = 97.94 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 86.47=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{86.47\times 2}{9.81}}\\\Rightarrow t=4.2\ s$

Time taken by the stone to travel 86.47 m to the water is is 4.2 seconds

The stone reaches the water after 4.2+1.52 = 5.72 seconds after throwing the stone

$v=u+at\\\Rightarrow v=0+9.81\times 86.47 = 848.27\ m/s$

Speed just before hitting the water is 848.27 m/s

3 0

2 years ago