answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Hitman42 [59]
2 years ago
5

The length of a wooden rod is 25.5 cm. What is this length in:(a) millimetres?(b) metres?​

Physics
2 answers:
Norma-Jean [14]2 years ago
8 0

Answer:

A. 255

B. 0.255

Explanation:

10mm = 1cm

100cm = 1m

<a href="https://en.wikiafripedia.org/wiki/Centimetre">Learn more</a>

https://en.wikiafripedia.org/wiki/Centimetre

larisa86 [58]2 years ago
4 0

Answer:

(a)255mm \\ (b)0.255m

Explanation:

(a)1cm = 10mm \\ 25.5cm = x

Cross multiply

1x = 25.5 \times 10 \\ x = 255mm

(b)100cm = 1m \\ 25.5cm = x

cross multiply

100x = 25.5 \\  \frac{100x}{100}  =  \frac{25.5}{100}  \\ x = 0.255m

hope this helps

brainliest appreciated

good luck! have a nice day!

You might be interested in
A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 t. what is the maximum hall voltage
Alex787 [66]
The equation for Hall voltage Vh is:

Vh=v*B*w, where v is the velocity of the strip, B is the magnitude of the magnetic field, and w is the width of the strip. 

v=25 cm/s = 0.25 m/s
B=5.6 T
w= 1.2 mm = 0.0012 m

We input the numbers into the equation and get:

Vh= 0.25*5.6*0.0012 = 0.00168 V

The maximum Hall voltage is Vh= 0.00168 V.
4 0
2 years ago
A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull
Elden [556K]

Answer:

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E)  351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

  • work done = force x distance x cos θ
  • since the rope is parallel to the ramp the angle between the rope and

        the ramp θ will be 0

       work done = 72 x 5.2 x cos 0

       work done by the rope = 374.4 J

(B) calculate the work done by gravity

  • the work done by gravity = weight of carton x distance x cos θ
  • The weight of the carton = force exerted by the mass of the carton = m x g
  • the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

      work done by gravity = 128 x 5.2  x cos 120

      work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

  • work done by the normal force = force x distance x cos θ
  • the angle between the normal force and the ramp is 90 degrees

       

         work done by the normal force = Fn x distance x cos θ

         work done by the normal force = Fn x 5.2 x cos 90

         work done by the normal force = Fn x 5.2 x 0

         work done by the normal force = 0 J

(D)  what is the net work done ?

  • The net work done is the addition of the work done by the rope,       gravitational force and the normal force

     net work done = 374.4 - 332.8 + 0 =  41.6 J  

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

  • work done by the rope= force x distance x cos θ
  • the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp

work done = 72 x 5.2 x cos 20

work done = 351.8 J

5 0
2 years ago
The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:
ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0
2 years ago
Which of the following statements are true about an object in two-dimensional projectile motion with no air resistance? (There c
ki77a [65]

Answer:

The correct answers are

The following statements are true about an object in two-dimensional projectile motion with no air resistance

D) The speed of the object is zero at its highest point.

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

Explanation:

A) The speed of the object is constant but its velocity is not constant.

False the vertical velocity increases on descent

B) The acceleration of the object is constant but its object is + g when the object is rising and -g when it is falling.

False, the acceleration is -g when the object is rising

C) The acceleration of the object is zero at its highest point.

False, the acceleration is constant in magnitude throughout the motion

D) The speed of the object is zero at its highest point.

True, the direction of motion changes at the highest point from hence the body comes to rest and the speed is zero

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

True, the horizontal acceleration has associated force during motion but the vertical acceleration is due to gravity which is constant downwards

6 0
2 years ago
A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which
goldfiish [28.3K]

Answer:

160 Hz  ,  240 Hz  , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

f_1=\dfrac{320}{4}\ Hz

f₁=80 Hz

Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

Frequency of second harmonic = f₃

f₃= 3 f₁=3 x 80 = 240 Hz

Frequency of fifth harmonic = f₅

f₅=  5 f₁= 5 x 80 = 400 Hz

Three frequencies are as follows

160 Hz  ,  240 Hz  , 400 Hz

6 0
2 years ago
Other questions:
  • engineers who design battery operated devices suck as sell phones and MP3 players try to make them as efficient as possible. An
    5·1 answer
  • A spherical balloon is 40 ft in diameter and surrounded by air at 60°F and 29.92 in Hg abs.(a) If the balloon is filled with hyd
    6·2 answers
  • A person in a boat sees a fish in the water (n = 1.33, the light rays making an angle of 40? relative to the water's surface. wh
    6·1 answer
  • A 2 kg bar of metal weighs about 4.4 pounds. Approximately how much does it weigh in newtons?
    10·2 answers
  • A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T.
    5·2 answers
  • A substance has a heat of vaporization of 16.69 kJ/mole. At 254.3 K it has a vaporpressure of 92.44 mm Hg. Calculate its vapor p
    8·1 answer
  • For tax and accounting purposes, corporations depreciate the value of equipment each year. One method used is called "linear dep
    10·1 answer
  • A girl is shown at position A on a swing when the seat is directly below the support bar. The seat is then at height A as shown
    7·2 answers
  • Un tren parte de la ciudad A, a las 8 h. con una velocidad de 50 km/h, para llegar a la ciudad B a las 10 h. Allí permanece dura
    10·1 answer
  • Select all the correct answers.
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!