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In-s [12.5K]
3 years ago
5

Dale (m = 75 kg) skis across a horizontal surface at a velocity of 5 m/s to the west. He drags his ski

Physics
1 answer:
masya89 [10]3 years ago
5 0

Answer:

1 m/s west

Explanation:

Impulse = change in momentum

J = m Δv

Taking east to be positive:

300 kg m/s = (75 kg) (v − (-5 m/s))

v = -1 m/s

His new velocity is 1 m/s west.

You might be interested in
What is tensor quantity?<br>Is Inertia a tensor? give reason​
pickupchik [31]

Answer:

A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts. Stress and strain are both tensor quantities. ... A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts.

Inertia Tensor. where I = the inertia tensor. The angular momentum of a rigid body rotating about an axis passing through the origin of the local reference frame is in fact the product of the inertia tensor of the object and the angular velocity. ... As shown in [7], the inertia tensor is symmetric.

Explanation:

Hope dis help

5 0
3 years ago
What is the average velocity of atoms in 1.00 mol of neon (a monatomic gas) at 465 K? For m, use 0.0202 kg.
TiliK225 [7]

Answer: 757m/s

Explanation:

Given the following :

Mole of neon gas = 1.00 mol

Temperature = 465k

Mass = 0.0202kg

Using the ideal gas equation. For calculating the average kinetic energy molecule :

0.5(mv^2) = 3/2 nRt

Where ;

M = mass, V = volume. R = gas constant(8.31 jK-1 mol-1, t = temperature in Kelvin, n = number of moles

Plugging our values

0.5(0.0202 × v^2) = 3/2 (1 × 8.31 × 465)

0.0101 v^2 = 5796.225

v^2 = 5796.225 / 0.0101

v^2 = 573883.66

v = √573883.66

v = 757.55109m/s

v = 757m/s

5 0
3 years ago
A bullet is fired into the air at an angle of 45°. How far does it travel before it is 1,000 feet above the ground? (Assume that
Readme [11.4K]

Answer:

It travels 1414 feets.

Explanation:

Let's take the length the bullet travels <em>l </em>as the hypotenuse of a right triangle and the height it reaches one of its sides. Since we got the angle α at which it was fired and the height <em>h</em> it reached, we can calculate <em>l</em> using the <em>sin(α)</em> function:

sin(\alpha )=\frac{opposite side}{hypotenuse}\\sin(\alpha)=\frac{h}{l}\\l=\frac{h}{sin(\alpha)}

Replacing:

l=\frac{1000ft}{sin(\frac{\pi}{4})}

Solving and roundin to the nearest foot:

l=1414 ft

3 0
4 years ago
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m

So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

6 0
3 years ago
If anyone knows how to do any of these PLEASE help me....im am so confused rn and our teacher sucks at explaining this stuff....
Taya2010 [7]
Take 68.2/60 = 1.137 hr
take 56.9/1.137 = 50.043 mi/hr

take 189/211 = 0.896

24.8/2 = 12.4 m
12.4/82.3 = 0.15s

7 0
3 years ago
Read 2 more answers
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