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Vinvika [58]
3 years ago
9

Calculate the density of a tin of mass 100g whose dimensions are 2cmx5cmx​

Physics
1 answer:
Luda [366]3 years ago
7 0

Answer:

your question in not complete.

you need to the high too.

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A football is kicked straight up from a height of 5 feet with an initial speed of 55 feet per second. The formula h equals negat
makvit [3.9K]

Answer:

3.53 second

Explanation:

The formula for the height is

h=-16t^{2}+55t+5

When it hits the ground, the height is zero.

So, put h = 0 in the above equation

0=-16t^{2}+55t+5

16t^{2}-55t-5=0

t=\frac{+55\pm \sqrt{55^{2}+4\times 5\times 16}}{2\times 16}

t=\frac{+55\pm 57.84}{2\times 16}

Take positive sign

t = 3.53 second.

Thus, the time taken to hit the ground is 3.53 second.

8 0
3 years ago
The angle of projection of projectile at which the range and height of projectile are
kozerog [31]

Answer:

76

At what angle height and range of projectile are equal?

It means that the range of the projectile with a given initial velocity is same for a pair of projection angles θ and 90° – θ.

7 0
3 years ago
alayn h uderstands the half life of a certain substance is 90days. how long would it take for 80lbs of her substance to be reduc
yan [13]

Answer:

135 days...............

7 0
3 years ago
How much work does a student (m = 60 kg) do when he Climb a tower 80 m high?
gtnhenbr [62]

Answer:

W = 47040  J

Explanation:

Given that,

The mass of a student, m = 60 kg

Height of the tower, h = 80 m

We need to find the work done in climbing the tower. The work done is given by :

W = mgh

So,

W = 60 × 9.8 × 80

W = 47040  J

So, the required work done is 47040  J.

7 0
3 years ago
In a physics laboratory experiment, a coil with 250 turns enclosing an area of 10.5 cm2 is rotated during the time interval 3.10
olchik [2.2K]
I guess the problem is asking for the induced emf in the coil.

Faraday-Neumann-Lenz states that the induced emf in a coil is given by:
\epsilon = -N \frac{\Delta \Phi}{\Delta t}
where
N is the number of turns in the coil
\Delta \Phi is the variation of magnetic flux through the coil
\Delta t is the time interval

The coil is initially perpendicular to the Earth's magnetic field, so the initial flux through it is given by the product between the magnetic field strength and the area of the coil:
\Phi_i = BA=(5.30 \cdot 10^{-5}T)(10.5 \cdot 10^{-4} m^2)=5.57 \cdot 10^{-8} Wb
At the end of the time interval, the coil is parallel to the field, so the final flux is zero:
\Phi_f = 0

Therefore, we can calculate now the induced emf by using the first formula:
\epsilon = -N  \frac{\Delta \Phi}{\Delta t}=- (250)  \frac{5.57 \cdot 10^{-8} Wb - 0}{3.10 \cdot 10^{-2} s} = -4.5 \cdot 10^{-4} V
7 0
4 years ago
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