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3 years ago

How much force is required to accelerate a 50 kg mass at 2 m/s2?

Physics

1 answer:

skelet666 [1.2K]3 years ago

5 0

Newton's 2nd law of motion:

Force = (mass) x (acceleration)

= (50 kg) x (2 m/s²)

= 100 newtons .

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The mass of a sample of iron is 31.5 g. The volume is 5 cm3. What is the density of iron in g/cm3?

ELEN [110]

Answer : 6.3 g/cm3
Step by step explanation:
Density = mass/volume

3 0

2 years ago

When droplets of water in the atmosphere act like prisms, the colors in sunlight undergo?

dexar [7]

White light is all the colours of light combined. When the droplets act like prisms, they split the white light into all its colours and also slightly bend the different colours. This is how a rainbow is formed.

4 0

3 years ago

A fullback preparing to carry the football starts from rest and accelerates straight ahead. He is handed the ball just before he

RideAnS [48]

Answer:

x=4.06m

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2} (3)\\

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for this problem

Vf=7.6m/s

t=1.07

Vo=0

we can use the ecuation number one to find the acceleration

a=(Vf-Vo)/t

a=(7.6-0)/1.07=7.1m/s^2

then we can use the ecuation number 2 to find the distance

{Vf^{2}-Vo^2}/{2.a} =X

(7.6^2-0^2)/(2x7.1)=4.06m

4 0

3 years ago

An 88 kg worker stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale reads 900 N. Fi

Aleks [24]

Answer:

The magnitude is "3.8 m/s²", in the upward direction.

Explanation:

The given values are:

Mass,

m = 88 kg

Scale reads,

T = 900 N

As we know,

⇒ $N=mg$

On substituting the given values, we get

⇒ $=88\times 9.8$

⇒ $=862.4 \ N$

Now,

⇒ $T=mg-ma$

On substituting the given values in the above equation, we get

⇒ $900=862.4-9.8 a$

On subtracting "862.4" from both sides, we get

⇒ $900-862.4=862.4-9.8 a-862.4$

⇒ $37.6=-9.8a$

⇒ $a=-\frac{37.6}{9.8}$

⇒ $a=3.8 \ m/s^2$ (upward direction)

8 0

3 years ago

Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=

drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

$v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2$

$v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2$

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

$v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

$v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

The final relative velocity of the satellite, $v_f$ = v₁ + v₂

∴ $v_f$ = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, $v_f$ = 0.190 m/s

4 0

2 years ago