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2 years ago

a bee flies 25m north of the hive, then 10m east, 5m west and 10m south. how far north and east of the hive is it now?

Physics

2 answers:

Vlad [161]2 years ago

8 0

Answer: 15m North and 5m East

Explanation: The bee first flies 25m northwards from the hive, and then from that point, flies 10m eastwards after which he flies westwards by 5m. This equates to it flying just 5m east (since 10m east — 5m west = 5m east).

Therefore the bee is just 5m east to the position of the hive.

From this point, it flies back down for 10m (southwards). This means of the 25m covered to the north of the hive, it has taken off 10m from it (25m north — 10m south = 15m north)

Therefore the bee is just 15m north of the location of the hive.

To pinpoint its location at this point in time, we would say the bee is located 5m to the East and 15m to the North of the hive.

krok68 [10]2 years ago

7 0

(25m north) + (10m south) = 15m north

(10m east) + (5m west) = 5m east

So the bee is now 15m north and 5m east of the hive.

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An inline skater skates on a circular track 120.0 m in diameter at a tangential speed of 9.20 m/s. If the skater’s mass is 68.5

jok3333 [9.3K]

Answer:

The centripetal force acting on the skater is <u>48.32 N.</u>

Explanation:

Given:

Radius of circular track is, $R=120.0\ m$

Tangential speed of the skater is, $v=9.20\ m/s$

Mass of the skater is, $m=68.5\ kg$

We are asked to find the centripetal force acting on the skater.

We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.

Centripetal force acting on the skater is given as:

$F_c=\frac{mv^2}{R}$

Now, plug in the given values of the known quantities and solve for centripetal force, $F_c$ . This gives,

$F_c=\frac{68.5\times (9.20)^2}{120.0}\\\\F_c=\frac{68.5\times 84.64}{120}\\\\F_c=\frac{5797.84}{120}\\\\F_c=48.32\ N$

Therefore, the centripetal force acting on the skater is 48.32 N.

3 0

2 years ago

You slip a wrench over a bolt. Taking the origin at the bolt, the other end of the wrench is at x=18cm, y=5.5cm. You apply a for

mart [117]

Answer:

The torque on the wrench is 4.188 Nm

Explanation:

Let r = xi + yj where is the distance of the applied force to the origin.

Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,

r = 0.18i + 0.055j

The applied force f = 88i - 23j

The torque τ = r × F

So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j

= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j

= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0 since i × i = 0, j × j = 0, i × j = k and j × i = -k

= 0 - 4.14k + 0.0484(-k) + 0

= -4.14k - 0.0484k

= -4.1884k Nm

≅ -4.188k Nm

So, the torque on the wrench is 4.188 Nm

8 0

2 years ago

A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm

Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

$mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g$

The mass of the meter stick is 126.99115 g

6 0

2 years ago

Read 2 more answers

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 damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is app

Cerrena [4.2K]

Answer:

0.5% per oscillation

Explanation:

The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.

Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

W^2= k/m--------------------------------------------------------------------------------------(3)

Slotting values into equation (3).

= 10/2.5.

= ✓4.

= 2 s^-1.

Recall that, F= -kX

F^2= (-0.1)^2

Potential energy,PE= 1/2 ×0.01

Potential energy= 0.05 ×100

= 0.5% per oscillation.

6 0

2 years ago