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Sindrei [870]
4 years ago
15

Question 2 (1 point) The distance traveled by a race car is 7 km in 100 sec. What is its speed?

Physics
1 answer:
fiasKO [112]4 years ago
7 0

Answer:

70 m/s

Step by step explanation:

<em>Step 1: gather your basic facts</em>

distance: 7 km

time: 100 s

<em>Step 2: use the formula of speed</em>

speed=\frac{distance}{time}

s=\frac{7000 m}{100 s}=70 m/s

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A machine lifts a 35 kg object doing 6860 J worth of work. How much power is produced by the machine if it lifts the object in 4
timofeeve [1]

Answer:

Explanation:

We need the power equation for this which is

P = Work/time

We have everything we need to solve this (the mass of the object is extra information):

P = 6860/4

P = 1715W

5 0
3 years ago
What displacement do I have if I travel at 10 m/s E for 10 s? <br> I need this fast
lapo4ka [179]

Answer:

100m.

Explanation:

Simply put, displacement is how far you've been displaced from your starting location. say you start at 0. You now travel E (assuming East) at 10 m/s for 10 s. Now, let's assume m is meters and s is seconds. you moved east at 10 meters per second for 10 seconds. This means that for each second you moved east, you moved 10 meters. Therefore, your displacement is 100 meters because 10 times 10 is 100. So you would write 100m as your displacement.

6 0
3 years ago
A spring stretches 0.018 m when a 2.8-kg object is suspendedfrom its end. How much mass should be attached to this spring sothat
agasfer [191]

Answer:

m = 4.29 kg

Explanation:

Given that,

Mass of the object, m = 2.8 kg

Stretching in the spring, x = 0.018 m

Frequency of vibration, f = 3 Hz

Let m is the mass of the object that is attached to the spring. When it is attached the gravitational force is balanced by the force on spring. It is given by :

mg=kx

k=\dfrac{mg}{x}

k=\dfrac{2.8\times 9.8}{0.018}

k = 1524.44 N/m

Since, \omega=\sqrt{\dfrac{k}{m}}

m=\dfrac{k}{4\pi^2 f^2}

m=\dfrac{1524.44}{4\pi^2 \times 3^2}

m = 4.29 kg

So, the mass that is attached to this spring is 4.29 kg. Hence, this is the required solution.

4 0
3 years ago
A 1250-kg compact car is moving with velocity v1 =36.2i^+12.7j^m/s. It skids on a frictionless icy patch and collides with a 448
MA_775_DIABLO [31]

Momentum is conserved, so the sum of the separate momenta of the car and wagon is equal to the momentum of the combined system:

(1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s) = ((1250 + 448) kg) <em>v</em>

where <em>v</em> is the velocity of the system. Solve for <em>v</em> :

<em>v</em> = ((1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s)) / (1698 kg)

<em>v</em> ≈ (30.3 <em>i</em> + 12.0 <em>j</em> ) m/s

7 0
3 years ago
A machine is supplied energy at a rate of 4,000 W and does useful work at a rate of 3,760 W. What is the efficiency of the machi
Dmitry_Shevchenko [17]
                = (3,760 joule/sec) / (4,000 joule/sec)
 
               =    3,760 / 4,000  =  0.94  =  94%
8 0
4 years ago
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