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melamori03 [73]
3 years ago
10

A ball thrown horizontally from the top of a building hits the ground in 0.600 s. If it had been thrown with twice the speed in

the same direction, it would have hit the ground in:________.
a. 4.0 s.
b. 1.0 s.
c. 0.50 s.
d. 0.25 s.
e. 0.125 s.
Physics
1 answer:
ludmilkaskok [199]3 years ago
5 0

Answer:

none of the answers is correct, the time  is the same  t₁ = t₂ = 0.600 s

Explanation:

This is a kinematics exercise, analyze the situation a bit. The vertical speed in both cases is the same is zero, the horizontal speed in the second case is double (vₓ₂ = 2 vₓ₁)

let's find the time to hit the ground

     y = y₀ + I go t - ½ g t²

     0 = y₀ - ½ g t²

     t = √ 2y₀ / g

with the data from the first launch

     y₀i = ½ g t²

     y₀ = ½  9.8  0.6²

     y₀ = 1,764 m

with this is the same height the time to descend in the second case is the same

    t₂ = 0.600 s

this is because the horizontal velocity change changes the offset on the x axis, but does not affect the offset on the y axis

Therefore, none of the answers is correct, the time  is the same

t₁ = t₂ = 0.600 s

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3 years ago
A 13561 N car traveling at 51.1 km/h rounds
Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

a_{c} = \frac{v^{2}}{r}

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

F_{c} = ma_{c}

Where:

m: is the mass of the car

The mass is given by:

P = m*g

Where P is the weight of the car = 13561 N

m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg

Now, the centripetal force is:

F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

F_{c} = F_{\mu}

F_{c} = \mu N = \mu P

\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!                

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A negative charge of -0.550 m exerts an upward <a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="5868766e
almond37 [142]

Answer:

a. +10.9μC

b. 0.600N and downward

Explanation:

To determine the magnitude of the charge, we use the force rule that exist between two charges which us expressed as

F=(kq₁q₂)/r²

since q₁=-0.55μC and the force it applied on the charge above it is upward,we can conclude that the second charge is +ve, hence we calculate its magnitude as

q₂=Fr²/kq₁

q₂=(0.6N*0.3²)/(9*10⁹*0.55*10⁻⁶)

q₂=0.054/4950

q₂=1.09*10⁻⁵c

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