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Mademuasel [1]
3 years ago
6

A ball is fired at an angle of 45 degrees, the angle that yields the maximum range in the absence of air resistance. What is the

ratio of the ball's maximum height to its range?

Physics
2 answers:
Nesterboy [21]3 years ago
7 0
Look at the picture for the answer

OleMash [197]3 years ago
5 0

Answer: 0.361:1

Explanation:

Since Range is U^2sin2Φ/g

And maximum height = U^2 sin^2Φ/2g

For range sinceΦ= 45

Range = U^2sin90/g

Sin90=1

Range= U^2/g

For maximum height

U^2sin^2Φ/2g

U^2sin(45)^2/2g

0.723U^2/2g

Ratio of maximum height to range

0.723U^2/2g*g/U^2

0.723/2

0.361

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weqwewe [10]

Answer:c

Explanation:it picking up speed

5 0
3 years ago
Read 2 more answers
Help please physics !!
malfutka [58]

Answer:

Option A. 1 bar = 1 atm

Explanation:

Pressure has various units of measurement. Each unit of measurement can be converted to other units of measurement. For example:

1 atm = 1 bar

1 atm = 760 mmHg

1 atm = 760 torr

1 atm = 1×10⁵ N/m²

1 atm = 1×10⁵ Pa

With the above conversion scale we can convert from one unit to the other.

Considering the question given above, it is evident from the coversion scale illustrated above that only option A is correct.

Thus,

1 bar = 1 atm

5 0
2 years ago
Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t
stepan [7]
We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
F=ma
where F=-50 N is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
a= \frac{v_f^2-v_i^2}{2d}
where
v_f=0 is the final speed of the sled
v_i=5 m/s is the initial speed
d is the distance covered

By rearranging the equation, we find d:
d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m
3 0
2 years ago
Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.
lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = (6\times0.1)kg/min=0.6kg/min

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = \frac{dx}{dt}

So, the rate of change of salt in the tank can be given by the following equation,

1000\frac{dx}{dt} =0.6-6x

or, \int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt

or, T = 693.147 min      (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0
3 years ago
Ur a genius if u explain how it’s A correctly
Alinara [238K]

Answer:

Explanation:

4/1=4

3/2=1.5

2/3=0.666667

1/4=0.25

D has the least number so its D

6 0
2 years ago
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