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3 years ago

What is the momentum of 70 kg runner traveling at 10 m/s?

Physics

2 answers:

Ivenika [448]3 years ago

8 0

700 kg m/s
Because momentum is mass times velocity

valentina_108 [34]3 years ago

5 0

Answer : The momentum of 70 kg runner is, 700 kg.m/s

Explanation :

Momentum : It is defined as the motion of a moving body. Or it is defined as the product of mass of velocity of an object.

Formula of momentum is:

$p=mv$

where,

p = momentum = ?

m = mass = 70 kg

v = velocity = 10 m/s

Now put all the given values in the above formula, we get:

$p=(70kg)\times (10m/s)$

$p=700kg.m/s$

Therefore, the momentum of 70 kg runner is, 700 kg.m/s

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Suppose that a particular artillery piece has a range R = 9880 yards . Find its range in miles. Use the facts that 1mile=5280ft

Solnce55 [7]

Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

6 0

3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago

How long will it take for a car to travel 200 km if it has an average speed of 55 km/hr?

andreyandreev [35.5K]

Answer: Approximately 3.65 hours

Explanation:

55 km/h x 3.65 hrs = 200.75 Km/h

4 0

3 years ago

on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker

Aleksandr-060686 [28]

B4 the tackle:

The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north 

and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east 

After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle 

The vector triangle is right angled: 

magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s 

so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] 

v(f) = 5.6 m/s (to 2 sig figs) 

direction of v(f) is the same as the direction of the final momentum 

so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) 

so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E 

btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it

4 0

3 years ago

When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured

zaharov [31]

Answer:

$\gamma=6.07*10^5\frac{N}{m^2}$

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

$\gamma=\frac{F/A}{\Delta L/L_0}$

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, $\Delta L$ is the amount by which the length of the object changes and $L_0$ is the original length of the object. In this case the force is the weight of the mass:

$F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N$

Replacing the given values in Young's modulus formula:

$\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}$

6 0

3 years ago