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3 years ago

Mathphys please notice me

Physics

2 answers:

eimsori [14]3 years ago

7 0

Answer:

24.2 s

29.2 m

Explanation:

First, convert each speed from km/h to m/s.

91.5 km/h = 25.42 m/s

77.5 km/h = 21.53 m/s

The speed of the cheetah relative to the gazelle is:

v = 25.42 m/s − 21.53 m/s

v = 3.89 m/s

So the time it takes for the cheetah to catch the gazelle is:

x = v t

94.2 m = (3.89 m/s) t

t = 24.2 s

If instead we want to find x so that t = 7.5 s, then:

x = v t

x = (3.89 m/s) (7.5 s)

x = 29.2 m

Marianna [84]3 years ago

3 0

Answer:

24.2 s

29.2 m

Explanation:

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A Porsche 944 Turbo has a rated engine power of 217hp . 30% of the power is lost in the drive train, and 70% reaches the wheels.

scZoUnD [109]

Explanation:

(a) It is given that two-third of weight is over the drive wheels. So, mathematically, w = $\frac{2}{3}mg$ .

Hence, maximum force is expressed as follows.

$F_{max} = \mu_{s} \times w$

$m \times a_{max} = \mu_{s} (\frac{2}{3} mg)$

Hence, the maximum acceleration is calculated as follows.

$a_{max} = \frac{2}{3} \mu_{s} \times g$

= $\frac{2}{3} \times 1.00 \times 9.8 m/s^{2}$

= 6.53 $m/s^{2}$

Hence, the maximum acceleration of the Porsche on a concrete surface where μs = 1 is 6.53 $m/s^{2}$ .

(b) Since, 30% of the power is lost in the drive train. So, the new power is 70% of $P_{max}$ .

That is, new power = $0.7 \times P_{max}$

Now, the expression for power in terms of force and velocity is as follows.

P = $F_{max} \nu$

$0.7 P_{max} = ma_{max} \nu$

Therefore, speed of the Porsche at maximum power output is as follows.

$\nu = 0.7 \times \frac{P_{max}}{ma_{max}}$

= $0.7 \times \frac{217 hp \times \frac{746 W}{1 hp}}{1500 kg \times 6.53 m/s^{2}}$

= 11.568 m/s

= 11.57 m/s

Therefore, speed of the Porsche at maximum power output is 11.57 m/s.

(c) The time taken will be calculated as follows.

time = $\frac{\text{velocity}}{\text{acceleration}}$

= $\frac{11.57 m/s}{6.53 m/s^{2}}$

= 1.77 s

Therefore, the Porsche takes 1.77 sec until it reaches the maximum power output.

6 0

4 years ago

What is the net Force needed to get a 16 kg box moving 4 m/s^2?

igomit [66]

Answer:

The net force should be of a magnitude of 64 N

Explanation:

We use Newton's second Law for this:

$F_{net} = m\,*\,a$

which for our case gives:

$F_{net} = m\,*\,a=16\,(4)\,N= 64\,\,N$

3 0

3 years ago

A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both

Len [333]

Answer:

D. Nothing will happen; the seesaw will still be balanced.

Explanation:

D. Nothing will happen; the seesaw will still be balanced. Since both toruqes or momentums respect to the center have changed in the same amount (one-half their original distance) the seesaw will remain balanced, if the children change distance in a different amount then it will be out of balance

3 0

3 years ago

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ERCRSOVOS - Table tennis footwork that is useful in this moment to cover a large distance quickly (SCRAMBLED)PE)

nata0808 [166]

Answer:

it's CROSSOVER I knew it because I'm a table tennis player

4 0

3 years ago

An object is pulled with a 85 N force inclined at 27° along a horizontal surface ABC

OLEGan [10]

Answer:

m = 15.15 kg

Explanation:

Newton's Second Law of motion states that when an unbalanced force is applied on a body, an acceleration is produced in it in the direction of force. The component of force along the horizontal direction here, will be given by the Newton's Second Law as:

Fx = ma

F Cosθ = ma

where,

F = Magnitude of Force = 85 N

θ = Angle with horizontal = 27°

m = mass of object = ?

a = acceleration of object = 5 m/s²

Therefore,

85 N Cos 27° = m(5 m/s²)

m = 75.73 N/5 m/s²

6 0

3 years ago