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rosijanka [135]

3 years ago

8

A 1500 kg car traveling at 80.0 km/h comes to a screeching halt in a time of 4.00 seconds. Calculate the force of friction exper

ienced by the car.

1 answer:

amm18123 years ago

7 0

Answer:

F=m x a

(F is force ,M is mass and A is acceleration)

in thisncase the Mass is given but we need to find ou the acceleration

Formula for acceleration-

a=(v - u)/t

(v is final velocity , u is initiatal velocity and t is time)

a = (0 - 80)/4

a= -80/4

a= -20

By substituting the values-

F= m x a

F= 1500 x -20

F=-30000N

Thus the force acted is -30000N

hope this helps

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A bullet of mass m=26g is fired into a wooden block of mass M=4.7kg. The block is attached to a string of length 1.5m. The bulle

vodka [1.7K]

Answer:

u = 449 m/s

Explanation:

Given,

Mass of the bullet, m = 26 g

Mass of the wooden block,M = 4.7 Kg

height of the block,h = 0.31 m

initial speed of the block, u = ?

Using conservation of energy

$(M+ m)gh = \dfrac{1}{2}(M+m)v^2$

$gh = \dfrac{1}{2}v^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.81\times 0.31}$

v = 2.47 m/s

Now, using conservation of momentum to calculate the speed of the bullet.

m u + M u' = (M+m)v

m u = (M+m)v

0.026 x u = (4.7+0.026) x 2.47

u = 449 m/s

Hence, the speed of the bullet is equal to 449 m/s.

5 0

3 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

So

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

So

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

So

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

3 years ago

Plz help me. I need urgent Why a magnetic compass is used to determine direction? Why a nail gets magnetized when kept close to

docker41 [41]

Answer:

Compasses are mainly used in navigation to find direction on the earth. This works because the Earth itself has a magnetic field which is similar to that of a bar magnet. The compass needle aligns with the Earth's magnetic field direction and points north-south. Also, In a magnet all the domains are oriented in the same direction. In the case of a nail, the domains can be aligned in the same direction causing the nail to become magnetic. That is because if you hang a bar magnet from a thread, the north pole will point to magnetic north. When you bring one north pole close to another north pole they repel each other. You can feel the two magnets pushing each other apart.

4 0

3 years ago

A 100 kg individual consumes 1200 kcal of food energy a day. Calculate

IrinaK [193]

Answer:

(a) 5142.86 m

(b) 317.5 m/s

(c) 49.3 degree C

Explanation:

m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J

(a) Let the altitude be h

Q = m x g x h

504 x 10^4 = 100 x 9.8 x h

h = 5142.86 m

(b) Let v be the speed

Q = 1/2 m v^2

504 x 10^4 = 1/2 x 100 x v^2

v = 317.5 m/s

(c) The temperature of normal human body, T1 = 37 degree C

Let the final temperature is T2.

Q = m x c x (T2 - T1)

504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)

T2 = 49.3 degree C

4 0

3 years ago

The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:

kow [346]

Answer:

$\frac{d_{1}}{d_{2}}=0.36$

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

$\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]}$ (1)

So, the temperature of the first and the second star will be:

$T_{1}=3866.7 K$

$T_{2}=6444.4 K$

Now the relation between the absolute luminosity and apparent brightness is given:

$L=l\cdot 4\pi r^{2}$ (2)

Where:

L is the absolute luminosity
l is the apparent brightness
r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

$\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}$

So the relative distance between both stars will be:

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}}$ (3)

The Boltzmann Law says, $L=A\sigma T^{4}$ (4)

σ is the Boltzmann constant
A is the area
T is the temperature
L is the absolute luminosity

Let's put (4) in (3) for each star.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}$

As we know both stars have the same size we can canceled out the areas.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=0.36$

I hope it helps!

5 0

3 years ago