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2 years ago

With what force will a car hit a tree if the car has a mass of 3,415 kg and it is accelerating at

Physics

1 answer:

weeeeeb [17]2 years ago

3 0

Answer:

$F= 17,075\ N$

Explanation:

When the car is under an accelerating force and hits a tree, the instant force received by the tree is the same force that is accelerating the car.

The accelerating force can be calculated using Newton's second law:

$F=m\cdot a$

Where m is the mass of the car and a is the acceleration.

$F=3,415\ kg\cdot 5\ m/s^2$

$\boxed{F= 17,075\ N}$

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Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in

Karo-lina-s [1.5K]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

Explanation:

(a) Self inductance is calculated as;

$L = \frac{N^2 \mu_0 A}{l}$

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

$A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2$

$L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH$

(b) The energy stored in the inductor when 21 A current ;

$E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J$

$emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s$

3 0

2 years ago

HELPPP IM DESPERATE

damaskus [11]

the first one cuz I know

5 0

2 years ago

We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 128 ∘C. The gas expands and, in the process, a

o-na [289]

Answer:

The final temperature of the gas is 114.53°C.

Explanation:

Firstly, we calculate the change in internal energy, ΔU from the first law of thermodynamics:

ΔU=Q - W

ΔU = 1180 J - 2020 J = -840 J

Secondly, from the ideal gas law, we calculate the final temperature of the gas, using the change in internal energy:

$ΔU=\frac{3}{2} nRΔT$

$ΔU=\frac{3}{2} nR(T_{2} -T_{1} )$

Then we make the final temperature, T₂, subject of the formula:

$T_{2} =\frac{2ΔU}{3nR} +T_{1}$

$T_{2} =\frac{2(-840J)}{(3)(5)(8.314J/mol.K)} +128 deg.C$

$T_{2} =114.53 deg.C$

Therefore the final temperature of the gas, T₂, is 114.53°C.

7 0

2 years ago

A 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?

Marianna [84]

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

Given the following data;

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

P.E = 3430J

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

Workdone = 3430Nm

6 0

2 years ago

In an electrolytic cell, to which electrode will a positive ion migrate and undergo reduction? the anode, which is negatively ch

ahrayia [7]

The reaction of reduction always undergoes with the cathode, the positive ion will migrate towards the cathode with the negative charge whilst the anode always has oxidation reaction. These two types of reaction does not change.

4 0

2 years ago

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