All categories

3 years ago

PLEASE HELP SOON

Physics

2 answers:

Kitty [74]3 years ago

7 0

Answer:you in connections too?

Explanation:

Sergio [31]3 years ago

7 0

The mass of the car is 1633 kg.

<u>Explanation:</u>

Kinetic energy is the energy attained by the object when it is in motion. So the kinetic energy exhibited by any object is directly proportional to the mass of the object as well as the square of velocity of the object.

Thus as here the kinetic energy is given as $4.32 \times 10^{5}$ J and speed or velocity is given a 23 m/s, the mass (m) of the object can be found as

$\text {Kinetic Energy}=\frac{1}{2} \times \text { m } \times \text { speed }^{2}$

$4.32 \times 10^{5}=\frac{1}{2} \times m \times 23 \times 23$

Thus,

$\text {m}=\frac{4.32 \times 10^{5} \times 2}{23 \times 23} = 1633 \mathrm{kg}$

You might be interested in

An object moves 20 m east in 30 s and then returns to its starting point taking an additional 50 s. If west is chosen as the pos

user100 [1]

Answer:

v = 0.5 m/s.

Explanation:

Total distance, d = 20 m + 20 m = 40 m

Total time taken, t = 30 s + 50 s = 80 s

The average speed of an object is the total distance divided by time taken. So,

$v=\dfrac{40}{80}\\\\=0.5\ m/s$

So, the average speed of the object is 0.5 m/s.

7 0

2 years ago

An unusual lightning strike has a vertical portion with a current of –400 A downwards. The Earth’s magnetic field at that locati

marusya05 [52]

Answer:

The magnetic force is 0.3 N.

Explanation:

Given that,

Current in an unusual lightning strike, I = -400 A

The Earth’s magnetic field at that location is parallel to the ground and has a magnitude of 30 μT, $B=30\times 10^{-6}\ T$

We need to find the force exerted by the Earth’s magnetic field on the 25 m-long current.

The magnetic force is given by :

$F=ILB\\\\F=400\times 25\times 30\times 10^{-6}\\\\F=0.3\ N$

So, the magnetic force is 0.3 N.

8 0

3 years ago

A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same m

Nina [5.8K]

Answer: $0.05\ mm$

Explanation:

Given

Cross-sectional area of wire $A_1=4\ mm^2$

Extension of wire $\delta l=0.1\ mm$

Extension in a wire is given by

$\Rightarrow \delta l=\dfrac{FL}{AE}$

where, $E=\text{Youngs modulus}$

$\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)$

for same force, length and material

$\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)$

Divide (i) and (ii)

$\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm$

5 0

2 years ago

What would happen to the marbles speed if we had longer tables

Zanzabum

Answer:

The marble will start to decrease in speed

Explanation:

The longer the table is the more the marbles speed would decrease because marbles can only be fast for a certain amount of time

7 0

3 years ago

Read 2 more answers

A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are

Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, $\Sigma F_y$ = 0

The vertical component of the tension, $T_y$ , in each half of the rope is given as follows;

$T_y$ = T × sin(θ)

∴ $\Sigma F_y$ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0

2 years ago