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mr Goodwill [35]
2 years ago
11

A 1700kg rhino charges at a speed of 50.0km/h. what is the magnitude of the average force needed to bring the rhino to a stop in

0.50s?
Physics
1 answer:
stepladder [879]2 years ago
7 0
Impulse equals Change in Momentum
F = average applied force = to be determined
Δt = time during which the force is applied = 0.50 s
m = mass = 1,700 kg
Δp = change in momentum = to be determined
Δv = change in velocity = to be determined
v1 = initial velocity = 50.0 km/h = 50,000 m/h = 13.9 m/s
v2 = final velocity = 0.00 km/h = 0.00 m/s

F∙Δt = Δp
F∙Δt = m∙Δv
F∙Δt = m∙(v2 - v1)

F = m∙(v2 - v1) / Δt
F = 1,700 kg∙(0.00 m/s - 13.9 m/s) / 0.50 s
<span>F = -47,222 N The negative sign means that the force vector is </span>
<span>applied AGAINST the momentum vector of the rhinoceros.</span>
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A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration
wel

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

V=X/T

First stage:

T1=5s

v_{f}  =v_{o} - at

But, v_{f}  =0   (decelerates to rest)

then: a =v_{o} /t=0.3/5=0.06m/s^{2}

on the other hand:

x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m

X1=75cm

Second stage:

T2=5s

x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

8 0
2 years ago
Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n
Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

<u>FOR FRIEND ON MOON</u>:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

<u>FOR ME ON EARTH</u>:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

<u>Therefore, the friend on moon will be richer.</u>

7 0
2 years ago
A moving sidewalk has a velocity of 1.7m/s north. if a man walks 1.1m/s, how long does it take him to travel 15m north in relati
alexira [117]
I think the answer will be a
3 0
2 years ago
A uniform magnetic field makes an angle of 30o with the z axis. If the magnetic flux through a 1.0 m2 portion of the xy plane is
Irina18 [472]

Answer:

(b) 10 Wb

Explanation:

Given;

angle of inclination of magnetic field, θ = 30°

initial area of the plane, A₁ = 1 m²

initial magnetic flux through the plane, Φ₁ =  5.0 Wb

Magnetic flux is given as;

Φ = BACosθ

where;

B is the strength of magnetic field

A is the area of the plane

θ is the angle of inclination

Φ₁ = BA₁Cosθ

5 = B(1 x cos30)

B = 5/(cos30)

B = 5.7735 T

Now calculate the magnetic flux through a 2.0 m² portion of the same plane

Φ₂ = BA₂Cosθ

Φ₂ = 5.7735 x 2 x cos30

Φ₂ = 10 Wb

Therefore, the magnetic flux through a 2.0 m² portion of the same plane is is 10 Wb.

Option "b"

3 0
2 years ago
A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point
il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

T + mg =\frac{mv^{2}}{r}

Inserting the values

T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}

T = 31.1 N

7 0
2 years ago
Read 2 more answers
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