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poizon [28]
2 years ago
5

A car starts from rest and acquires a velocity of 50m/s in 3secs. Calculate i) acceleration ii) distance covered.

Physics
1 answer:
mafiozo [28]2 years ago
4 0

Answer: 75.02 m

Explanation:

u = 0 ( starts from rest )

v = 50 m/s

t = 3 s

( i ) a = v - u / t

= 50 - 0 /3

= 16.67

( ii ) s = ut + 1/2 at²

= 0 × 3 + 1/2 × 16.67 × 3 × 3

= <u>75.02 m</u>

Hope this helps...

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Select all of the following that describes momentum, p [mark all correct answers]
Inga [223]

Answer:

c. Momentum is the product of mass and velocity

e. Momentum is a vector quantity

g. Momentum has unit of kgm/s

Explanation:

Linear momentum P

P = m .v

m =mass

v=Velocity

If mass take in kg and velocity is in m/s then momentum p will be in kg.m/s.

1. momentum is the product of velocity and mass.

2.Momentum is a vector quantity.

3.Momentum have kg.m/s unit.

So the following option are correct.

c. Momentum is the product of mass and velocity

e. Momentum is a vector quantity

g. Momentum has unit of kgm/s.

Note-

1.Joule is the unit of energy.

2.One-half the product of mass and the square of the object's speed is known as kinetic energy.

7 0
3 years ago
Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attache
HACTEHA [7]

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f  is when n = 1. So,

f = 1/2L√(T/μ) =  1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f'  is when n = 3. So,

f' = 3/2L√(T/μ) =  3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

8 0
2 years ago
How do lines of latitude affect how direct or indirect the Sun’s rays are on the Earth?
IgorLugansk [536]
The technical definition of latitude is the angular distance north or south from the earth's equator measured through 90 degrees. ... Locations at lower latitudes receive stronger and more direct sunlight than locations near the poles. Energy input from the sun is the main driving force in the atmosphere.



The Seasons at Different Latitudes
The seasonal effects are different at different latitudes on Earth. Near the equator, for instance, all seasons are much the same. Every day of the year, the Sun is up half the time, so there are approximately 12 hours of sunshine and 12 hours of night.



When we consider Latitude alone as a control, we know that the low latitudes (say from the Equator to approximately 30 degrees N/S) are the warmest across the year (on an annual basis).
8 0
3 years ago
A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa
DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

3 0
3 years ago
Why is it important for a muscle to be attached to a fixed origin at one end and a moving insertion at the other? Discuss how th
vlabodo [156]
Muscles function only by contracting. This makes it necessary for one end of the muscle to be fixed and the other mobile.
Take the bicep for example.
Its origin is at the shoulder and its two heads connect to the bones of the forearm, the radius and ulna.
Now, had the muscle not been fixed at one end, and contracted, it would pull both our shoulder and forearm together resulting in an ineffective movement. The desired motion is to lift the forearm (proximal and distal movement) which can only be achieved if the bicep is fixed at the shoulder and allowed to move at the forearm.
6 0
2 years ago
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